Answer:
place value of 3 in 783.97 is 3
Step-by-step explanation:
Answer:
Units
Step-by-step explanation:
The units start counting from 3 because after the point that is the 9 start counting tenth
Given log32≈0.631 and log37≈1.771, what is log314
Answer:
the log to the base 3 of 14 is 2.402
Step-by-step explanation:
You must find a way to indicate that 3 is the base; you cannot run this '3' together with 2, 7 or 14.
Example:
log to the base 3 of 2 = 0.631
log to the base 3 of 7 = 1.771
Note that 2 times 7 is 14. Thus, to obtain the log to the base 3 of 14, we must ADD the two logs shown above:
0.631
+1.771
----------
2.402
Thus, the log to the base 3 of 14 is 2.402.
Check: Does 3^2.402 = 14? YES
Sodas in a can are supposed to contain an average 12 oz. This particular brand has a standard deviation of 0.1 oz, with an average of 12.1 oz. If the can’s contents follow a Normal distribution, what is the probability that the mean contents of a six-pack are less than 12 oz?
Answer:
The probability is [tex]P(X < 12) = 0.99286[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 12 \ oz[/tex]
The standard deviation is [tex]\sigma = 0.1 \ oz[/tex]
The sample mean is [tex]\= x = 12.1 \ oz[/tex]
The sample size is n = 6 packs
The standard error of the mean is mathematically represented as
[tex]\sigma_{\= x } = \frac{\sigma}{\sqrt{n} }[/tex]
substituting values
[tex]\sigma_{\= x } = \frac{0.1}{\sqrt{6} }[/tex]
[tex]\sigma_{\= x } = 0.0408[/tex]
Given that the can’s contents follow a Normal distribution then then the probability that the mean contents of a six-pack are less than 12 oz is mathematically represented as
[tex]P(X < 12) = P ( \frac{X - \mu }{ \sigma_{\= x }} < \frac{\= x - \mu }{ \sigma_{\= x }} )[/tex]
Generally [tex]\frac{X - \mu }{ \sigma_{ \= x }} = Z (The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(X < 12) = P ( Z < \frac{\= x - \mu }{ \sigma_{\= x }} )[/tex]
substituting values
[tex]P(X < 12) = P ( Z < \frac{12.2 -12 }{0.0408} )[/tex]
[tex]P(X < 12) = P ( Z < 2.45 )[/tex]
From the normal distribution table the value of [tex]P ( Z < 2.45 )[/tex] is
[tex]P (Z < 2.45)0.99286[/tex]
=> [tex]P(X < 12) = 0.99286[/tex]
Study the table. Which best describes the function represented by the data in the table?
Answer:
linear with a common first difference of 2
Step-by-step explanation:
On the face of it, you can reject answers that ascribe a common ratio to a linear or quadratic function. (A common ratio is characteristic of an exponential function.)
You can also reject the answer that ascribes a common first difference to a quadratic function. (A quadratic function has a common second difference.)
After you reject the nonsense answers, there is only one remaining choice. It is also the correct one:
linear with a common first difference of 2
_____
The ratio of change in y to change in x is ...
(0 -(-2))/(-2 -(-3)) = 2
(4 -0)/(0 -(-2)) = 2
(12 -4)/(4 -0) = 2
That is, y increases by 2 when x increases by 1. The common first difference is 2.
Four couples are at a party. Four of the eight people are randomly selected to win a prize. No person can win more than one prize. What is the probability that both of the members of at least one couple win prizes? Express your answer as common fraction.
Answer:
27/35
Step-by-step explanation:
We use combination to solve for this
C(n, r), =nCr = n!/r!(n - r)!
From the question, we are told that:
Four couples are at a party. Four of the eight people are randomly selected to win a prize.
Four couples = 8 people.
= 8C4 = 8!/4! (8 - 4)!
= 70
No person can win more than one prize. ( No person can win more than one prize of the 4 people selected)
This can happen in 4 ways
[4C1 × 3C2 ] × 4=
[4!/1! ×( 4 - 1)!] × [3!/2! ×(3-2)!] × 4 ways
= 4 × 3 × 4 ways
= 48
The probability that both of the members of at least one couple win prizes
48 + 4C2/ 8C4
4C2 = 4!/2!(4 - 2) !
= 6
8C4 = 8C4 = 8!/4! (8 - 4)!
= 70
48 + 6/ 70
= 54/70
= 27/35
Therefore, the probability that both of the members of at least one couple win prizes is 27/35.
The probability that both of the members of at least one couple win prizes is 27/35 and this can be determined by using the given data.
Given :
Four couples are at a party. Four of the eight people are randomly selected to win a prize. No person can win more than one prize.The following steps can be used in order to determine the probability that both of the members of at least one couple win prizes:
Step 1 - The concept of probability is used in order to determine the probability that both of the members of at least one couple win prizes.
Step 2 - According to the given data, the total number of people is 8.
Step 3 - So, the probability that both of the members of at least one couple win prizes is:
[tex]\rm P =\dfrac{ \;^4C_1\times \;^3C_2\times 4 + \;^4C_2}{\;^8C_4}[/tex]
Step 4 - Simplify the above expression.
[tex]\rm P =\dfrac{48+ 6}{70}[/tex]
[tex]\rm P = \dfrac{27}{35}[/tex]
So, the probability that both of the members of at least one couple win prizes is 27/35.
For more information, refer to the link given below:
https://brainly.com/question/795909
Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x→9 x − 9 x2 − 81
Without resorting to L'Hopitâl's rule,
[tex]\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac{x-9}{(x-9)(x+9)}=\lim_{x\to9}\frac1{x+9}=\frac1{18}[/tex]
With the rule, we get the same result:
[tex]\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac1{2x}=\frac1{18}[/tex]
A survey showed that among 785 randomly selected subjects who completed four years of college, 144 of them are smokers and 84 do not smoke (based on data from the American Medical Association). Suppose you want to test at the 0.01 significance level the claim that the rate of smoking among those with four years of college is less than the 27% rate for the general population.
A. State the null and alternative hypotheses.
B. Find the sample statistic and the p-value.
C. What is your conclusion?
Answer:
We conclude that the rate of smoking among those with four years of college is less than the 27% rate for the general population.
Step-by-step explanation:
We are given that a survey showed that among 785 randomly selected subjects who completed four years of college, 144 of them are smokers.
Let p = population proportion of smokers among those with four years of college
So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\geq[/tex] 27% {means that the rate of smoking among those with four years of college is more than or equal to the 27% rate for the general population}
Alternate Hypothesis, [tex]H_A[/tex] : p < 27% {means that the rate of smoking among those with four years of college is less than the 27% rate for the general population}
The test statistics that will be used here is One-sample z-test for proportions;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of smokers = [tex]\frac{144}{785}[/tex] = 0.18
n = sample of subjects = 785
So, the test statistics = [tex]\frac{0.18-0.27}{\sqrt{\frac{0.27(1-0.27)}{785} } }[/tex]
= -5.68
The value of z-test statistics is -5.68.
Also, the P-value of the test statistics is given by;P-value = P(Z < -5.68) = Less than 0.0001
Now, at a 0.01 level of significance, the z table gives a critical value of -2.3262 for the left-tailed test.
Since the value of our test statistics is less than the critical value of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the rate of smoking among those with four years of college is less than the 27% rate for the general population.
Express the product of z1 and z2 in standard form given that [tex]z_{1} = 6[cos(\frac{2\pi }{5}) + isin(\frac{2\pi }{5})][/tex] and [tex]z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2}) + isin(\frac{-\pi }{2})][/tex]
Answer:
Solution : 5.244 - 16.140i
Step-by-step explanation:
If we want to express the two as a product, we would have the following expression.
[tex]-6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi }{5}\right)\right]\cdot 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right][/tex]
Now we have two trivial identities that we can apply here,
( 1 ) cos(- π / 2) = 0,
( 2 ) sin(- π / 2) = - 1
Substituting them,
= [tex]-6\cdot \:2\sqrt{2}\left(0-i\right)\left(\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi }{5}\right)\right)[/tex]
= [tex]-12\sqrt{2}\sin \left(\frac{2\pi }{5}\right)+12\sqrt{2}\cos \left(\frac{2\pi }{5}\right)i[/tex]
Again we have another two identities we can apply,
( 1 ) sin(x) = cos(π / 2 - x )
( 2 ) cos(x) = sin(π / 2 - x )
[tex]\sin \left(\frac{2\pi }{5}\right)=\cos \left(\frac{\pi }{2}-\frac{2\pi }{5}\right) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}[/tex]
[tex]\cos \left(\frac{2\pi }{5}\right)=\sin \left(\frac{\pi }{2}-\frac{2\pi }{5}\right) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}[/tex]
Substitute,
[tex]-12\sqrt{2}(\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}) + 12\sqrt{2}(\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4})[/tex]
= [tex]-6\sqrt{5+\sqrt{5}}+6\sqrt{3-\sqrt{5}} i[/tex]
= [tex]-16.13996 + 5.24419i[/tex]
= [tex]5.24419i - 16.13996[/tex]
As you can see option d is the correct answer. 5.24419 is rounded to 5.244, and 16.13996 is rounded to 16.14.
PLS HELP :Find all the missing elements:
Answer:
[tex]\large \boxed{\mathrm{34.2}}[/tex]
Step-by-step explanation:
[tex]\sf B= arcsin (\frac{b \times sin(A)}{a} )[/tex]
[tex]\sf B= arcsin (\frac{7 \times sin(40\°)}{8} )[/tex]
[tex]\sf B = 0.59733 \ rad = 34.225\°[/tex]
how many quarts are there in 12 gallons and 3 quarts? enter the number only. Do not include units
Answer:
51
Step-by-step explanation:
Simplify -2x^3 y x xy^2
Answer:
(4,4)⋅(4,4)
Step-by-step explanation:
The check_time function checks for the time format of a 12-hour clock, as follows: the hour is between 1 and 12, with no leading zero, followed by a colon, then minutes between 00 and 59, then an optional space, and then AM or PM, in upper or lower case. Fill in the regular expression to do that. How many of the concepts that you just learned can you use here
Answer:
Following are the correct code to this question:
import re#import package for regular expression
def check_time(text):#defining a method check_time that accepts string value
p = r'(1[012]|[1-9]):[0-5][0-9][ ]{0,1}?(am|pm|AM|PM)'#defining string variable p that stores values
val = re.search(p, text)#defining val variable that check serachs p and text variable values
return val!= None#use return keyword to return val value
print(check_time("12:45pm"))#defining print method that calls method by input value
print(check_time("9:59 AM")) #defining print method that calls method by input value
print(check_time("6:60 am")) #defining print method that calls method by input value
print(check_time("five o'clock"))#defining print method that calls method by input value
Output:
True
True
False
False
Step-by-step explanation:
In the above-given program, some data is missing that is code file so, the correct code can be defined as follows:
In the above-given method, that is "check time" it uses 12-hour time format validation, that is tested by coding the regex and all the value validates in the "val" variables, that can be defined as follows:
In the first step, its values should be in 1,2,3, ... 10,11,12 In the second step, it values in Between hour and minutes, and there will be a colon. In the third step, the minutes variable should take the double-digit, that will be like 00,01 .... 59. In the last step, one space becomes permitted after an hour: a minute or no space for am or pm value.4
If Randy flips a coin 3 times, what is the probability that it will come up heads 3 times?
Hi there! :)
Answer:
[tex]P(heads) = \frac{1}{8}[/tex]
Step-by-step explanation:
Probability of a coin landing on heads:
[tex]P(heads) = \frac{1}{2}[/tex]
Find the probability of getting heads 3 times:
[tex]\frac{1}{2} * \frac{1}{2} * \frac{1}{2} = \frac{1}{8}[/tex]
Therefore, the probability of the coin showing heads for 3 tosses is:
[tex]P(heads) = \frac{1}{8}[/tex]
g When conducting a one-way ANOVA, the _______ the between-treatment variability is when compared to the within-treatment variability, the __________the value of the F statistic will be which gives us ________ evidence against the null. (Choose all that apply)
Answer:
One - way ANOVA, the smaller the between treatment
The smaller the value of F statistic will give us significant evidence.
Step-by-step explanation:
ANOVA is a statistical technique designed to test mean of one or more quantitative populations. In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA. It is a statistical technique designed to test mean of one or more quantitative populations. In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA.
One-way ANOVA, the smaller the between treatment
The smaller the value of F statistic will give us significant evidence.
What is ANOVA?It should be the statistical technique that are made for testing the mean for one or more than one quantitative population. In two-way ANOVA it should be equivalent to the block mean. Here the column block represent the square be the three-way ANOVA.
Therefore, One-way ANOVA, the smaller the between treatment
The smaller the value of F statistic will give us significant evidence.
Learn more about evidence here: https://brainly.com/question/6764645?referrer=searchResults
The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft2). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y).
Temp. 174 176 177 178 178 179 180 181
Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74
Temp. 184 184 184 184 184 185 185 186
Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94
Temp. 186 186 186 188 188 189 190 192
Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16
(a) Determine the equation of the estimated regression line. (Round all numerical values to five decimal places.)
y =
(b) Calculate a point estimate for true average efficiency ratio when tank temperature is 186. (Round your answer to four decimal places.)
(c) Calculate the values of the residuals from the least squares line for the four observations for which temperature is 186. (Round your answers to four decimal places.)
(186, 0.94)
(186, 1.85)
(186, 2.02)
(186, 2.64)
(d) What proportion of the observed variation in efficiency ratio can be attributed to the simple linear regression relationship between the two variables? (Round your answer to four decimal places.)
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
Temp. 174 176 177 178 178 179 180 181
Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74
Temp. 184 184 184 184 184 185 185 186
Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94
Temp. 186 186 186 188 188 189 190 192
Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16
A)
Using the online linear regression calculator, the lie of best fit which models the data above is :
ŷ = 0.09386X - 15.55523
Where ;
X = independent variable
ŷ = predicted or dependent variable
- 15.55523 = intercept
0.09386 = gradient / slope
B)
Point estimate when tank temperature is 186
ŷ = 0.09386(186) - 15.55523
ŷ = 17.45796 - 15.55523
ŷ = 1.90273
C)
Residual error (y - ŷ), ŷ = 1.90273 when x = 186
(0.94 - 1.90273) = −0.96273
(1.85 - 1.90273) = −0.05273
(2.02 - 1.90273) = 0.11727
(2.64 - 1.90273) = 0.73727
D)
To determine the proportion of observed variation in efficiency ratio, we find the Coefficient of determination R^2, which can be found using the online Coefficient of determination calculator : the r^2 value obtained is 0.4433.
The following data represents the age of 30 lottery winners.
22 26 27 27 31 34
36 42 43 44 48 49
52 53 55 56 57 60
65 65 66 67 69 72
75 77 78 78 79 87
Complete the frequency distribution for the data.
Age Frequency
20-29
30-39
40-49
50-59
60-69
70-79
80-89
Answer:
Step-by-step explanation:
This is an example of a frequency distribution for a class interval. In order to complete the frequency distribution, we will count the number of data occurring in each group, and write that number as the frequency for that group. This is done as shown below:
Age Frequency ages in class
20-29 4 22, 26, 27, 27
30-39 3 31, 34, 36
40-49 5 42, 43, 44, 48, 49
50-59 5 52, 53, 55, 56, 57
60-69 6 60, 56, 65, 66, 67, 69
70-79 6 72, 75, 77, 78, 78, 79
80-89 1 87
Total 30
PLEAS HELP...FIRST CORRECT ANSWER WILL GET BRAINLIEST....PLEASE ANSWER NOW!!!!
The bar graph shows the number of students who earned each letter grade on an
exam, which statement about the graph is true?
*a clearer picture containing the graph is shown in the attachment
Answer:
20% of the class earned a D
Step-by-step Explanation:
Step 1: Determine the total number of students represented on the graph:
9 students => D
5 students => C
14 students => B
17 students => A
Total number of students = 45
Step 2: Express each category of students who scored a particular grade as a fraction and as percentage.
9 students => D => [tex] \frac{9}{45} = \frac{1}{5} [/tex] => as percentage, we have [tex] \frac{1}{5} * 100 = 20 percent [/tex]
5 students => C => [tex] \frac{5}{45} = \frac{1}{9} [/tex] => as percentage, we have [tex] \frac{1}{9} * 100 = 11.1 percent [/tex]
14 students => B => [tex] \frac{14}{45} [/tex] => as percentage, we have [tex] \frac{14}{45} * 100 = 31.1 percent [/tex]
17 students => A => [tex] \frac{17}{45} [/tex] => as percentage, we have [tex] \frac{17}{45} * 100 = 37.8 percent [/tex]
Step 3: Check each statement to see if they are true or not based on the calculations above.
Statement 1: "⅕ of the students earned a C."
This is NOT TRUE From our calculation, ⅑ of the students earned a C.
Statement 2: "3% more students earned an A than a B." This is also NOT TRUE.
37.8% earned A, while 31.1% earned a B. Thus, about 6.7% more students earned an A than a B.
Statement 3: "20% of the class earned a D". This is TRUE.
Check calculation in step 2.
Statement 4: "¼ of the class earned a B". This is NOT TRUE.
¼ is 25% of the class. Those who earned a B account for 31.1% not 25% (¼ of the class).
The correct statement is: "20% of the class earned a D"
How many pencils are in a bundle of 10
if they're in a bundle of 10 then theres 10 pencils
The denominator of a fraction is 30 more than the numerator. The value of the fraction is 3/5. Find the fraction.
Answer:
45
------
75
Step-by-step explanation:
Let x be the value of the numerator and x+30 be the value of the denominator
This is equal to 3/5
x 3
-------- = -------
x+30 5
Using cross products
5x = 3(x+30)
Distribute
5x = 3x+90
Subtract 3x from each side
2x = 90
Divide by 2
x = 45
The fraction is
45
-----
30+45
45
------
75
[tex]\dfrac{x}{x+30}=\dfrac{3}{5}\\\\5x=3(x+30)\\5x=3x+90\\2x=90\\x=45\\\\\dfrac{x}{x+30}=\dfrac{45}{45+30}=\dfrac{45}{75}[/tex]
i need help asap please
Answer:
[tex]x = -\frac{3}{2}[/tex] or [tex]x = 1[/tex]
Step-by-step explanation:
Using the zero product property, first step is to set the given equation, [tex] 2x^2 + x - 1 = 2 [/tex] , to zero. Then factorise the left side.
Thus,
[tex] 2x^2 + x - 1 = 2 [/tex]
Subtract 2 from both sides
[tex] 2x^2 + x - 1 - 2 = 2 - 2 [/tex]
[tex] 2x^2 + x - 3 = 0 [/tex]
Factorise the left side
[tex] 2x^2 + 3x - 2x - 3 = 0 [/tex]
[tex] x(2x + 3) - 1(2x + 3) = 0 [/tex]
[tex] (x - 1)(2x + 3) = 0 [/tex]
Find the solution
[tex] x - 1 = 0 [/tex]
Or
[tex]2x + 3 = 0[/tex]
[tex] x = 1 [/tex]
Or
[tex]2x + 3 = 0[/tex]
[tex]2x = -3[/tex]
[tex]x = -\frac{3}{2}[/tex]
The answer is: [tex] x = 1 [/tex] or [tex]x = -\frac{3}{2}[/tex]
what is the number if 4 is subtracted from the sum of one fourth of 5 times of 8 and 10
Answer:
Step-by-step explanation:
Lets, turn this into words and use order of operations, First, we look for multiplication and division.
the sum of one fourth of 5 times of 8 and 10 gets you 1/4(5*8) + 10 = 20
what is the number if 4 is subtracted from the sum
20 - 4 = 16
A sample of 81 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean.
Answer:
38.911≤p≤41.089
Step-by-step explanation:
The formula for calculating confidence interval for a population mean us as shown below;
CI = xbar ± Z×S/√N where;
xbar is the sample mean = 40
Z is the z score at 95% confidence interval = 1.96
S is the standard deviation = 5
N is the sample size = 81
Substituting this parameters in the formula we have;
CI = 40±1.96×5/√81
CI = 40±(1.96×5/9)
CI = 40±(1.96×0.556)
CI = 40±1.089
CI = (40-1.089, 40+1.089)
CI = (38.911, 41.089)
The 95% confidence interval for the population mean is 38.911≤p≤41.089
Answer:
38.9 ≤ U ≤ 41.1
Step-by-step explanation:
Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95
N = 81
The standard error, α(m) = α/√(N) = 5/√81 =5/9
Using table: 0.95 = 0.0379
Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96
Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}
But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1
(40 - 1.1 ≤ U ≤ 40 + 1.1)
∴ the confidence interval = 38.9 ≤ U ≤ 41.1
What value of x makes this equation true?
17 5 - 7 = -4
x=
y Su
What value of x makes this equation true? X/6-7=-4
Answer:
x=18
Step-by-step explanation:
x/6 - 7 = -4
x/6 = 3
(x/ 6) * 6 = 3*6
x = 18
Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n = 50 p = 0.2
Answer:
The mean, variance, and standard deviation of the binomial distribution are 10, 8, and 2.83 respectively.
Step-by-step explanation:
We have to find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p, i.e; n = 50 p = 0.2.
Let X = binomial random variable
So, X ~ Binom(n = 50, p = 0.2)
Now, the mean of the binomial distribution is given by;
Mean of X, E(X) = n [tex]\times[/tex] p
= 50 [tex]\times[/tex] 0.2 = 10
Now, the variance of the binomial distribution is given by;
Variance of X, V(X) = n [tex]\times[/tex] p [tex]\times[/tex] (1 - p)
= 50 [tex]\times[/tex] 0.2 [tex]\times[/tex] (1 - 0.2)
= 10 [tex]\times[/tex] 0.8 = 8
Also, the standard deviation of the binomial distribution is given by;
Standard deviation of X, S.D.(X) = [tex]\sqrt{\text{n} \times \text{p} \times (1 - \text{p})}[/tex]
= [tex]\sqrt{\text{50} \times \text{0.2} \times (1 - \text{0.2})}[/tex]
= [tex]\sqrt{8}[/tex] = 2.83
What is the error in this problem
Answer:
The angle of elevation is not properly represented
Step-by-step explanation:
Given
The question in the attachment;
Required
Determine the error
See attachment for proper representation of the angle of elevation;
Solving further (From the Attachment)
[tex]Tan22 = \frac{x}{3000}[/tex]
Multiply both sides by 3000
[tex]x = 3000 * tan22[/tex]
[tex]x = 3000 * 0.4040[/tex]
[tex]x = 1212[/tex]
The cliff is about 1212 feet high
plzzz help me quick will give goood rate
Answer:
Average rate of change of the function will be = (-1.5)
Step-by-step explanation:
Average rate of change of a function f(x) is determined by the formula,
Average rate of change = [tex]\frac{f(b)-f(a)}{b-a}[/tex] If a < x < b
We have to find the average rate of change of a function g(t) between the interval [-3, 1]
From the given table,
For t = -3,
g(-3) = 6
For t = 1,
g(1) = 0
Therefore, average rate of change of the function in the given interval
= [tex]\frac{g(1)-g(-3)}{1-(-3)}[/tex]
= [tex]\frac{0-6}{1+3}[/tex]
= [tex]-\frac{3}{2}[/tex]
= - 1.5
At Jefferson Middle School, eighty-two students were asked which sports they plan to participate in for the coming year. Twenty students plan to participate in track and cross country; six students in cross country and basketball; and eight students in track and basketball. Twelve students plan to participate in all three sports. A total of thirty students plan to participate in basketball, and a total of forty students plan to participate in cross country. Ten students don't plan to participate in any of the three sports. How many students plan to just participate in cross country? 2 4 40 30
Answer:
40
Step-by-step explanation:
In the question only lies the answer:
"and a total of forty students plan to participate in cross country."
Answer:
2
Step-by-step explanation:
2
Determine the point estimate of the population proportion and the margin of error for the following confidence interval.Lower boundequals0.212,upper boundequals0.758,nequals1200The point estimate of the population proportion is . 485.(Round to the nearest thousandth as needed.)The margin of error is 0.273.(Round to the nearest thousandth as needed.)
Answer: The point estimate of the population proportion is . 485.
The margin of error is 0.273.
Step-by-step explanation:
Confidence interval for population proportion(p):
sample proportion ± Margin of error
Given: Lower bound of confidence interval = 0.212
Upper bound = 0.758
⇒sample proportion - Margin of error=0.212 (i)
sample proportion + Margin of error= 0.758 (ii)
Adding (i) and (ii) , we get
2(sample proportion) =0.970
⇒ sample proportion = 0.970÷2= 0.485
Since sample proportion is the point estimate of the population proportion.
So, the point estimate of the population proportion= 0.485
Now put sample proportion =0.485 in (ii), we get
0.485+ Margin of error= 0.758
⇒ Margin of error= 0.758 - 0.485 =0.273
i.e. The margin of error is 0.273.
1+2x=6x+11 PLS HELP URGENT
Answer:
x = -5/2
Step-by-step explanation:
1+2x=6x+11
Subtract 2x from each side
1+2x-2x=6x-2x+11
1 = 4x+11
Subtract 11 from each side
1-11 = 4x
-10 =4x
Divide by 4
-10/4 = 4x/4
-5/2 =x
Answer:
[tex]\boxed{x=-\frac{5}{2}}\\[/tex]
Step-by-step explanation:
To begin, get the variable on one side of the equation - preferably the left for standard solution notation (for this equation, it is easier to place it on the right side to avoid negative values). Do this by subtracting 2x from both sides of the equation. Then, subtract 11. Finally, divide by 4 and get the answer in terms of x.
1 + 2x = 6x + 11
1 = 4x + 11
-10 = 4x
[tex]\boxed{x=-\frac{5}{2}}[/tex]
As a bowling instructor, you calculate your students' averages during tournaments. In 5 games, one bowler had the following scores: 143, 156, 172, 133, and 167. What was that bowler's average?
Answer:
154.2
Step-by-step explanation:
143 plus
156 plus
172 plus
133 plus
167 = 771
divide by 5 equals 154.2
In the Cash Now lottery game there are 8 finalists who submitted entry tickets on time. From these 8 tickets, three grand prize winners will be drawn. The first prize is one million dollars, the second prize is one hundred thousand dollars, and the third prize is ten thousand dollars. Determine the total number of different ways in which the winners can be drawn. (Assume that the tickets are not replaced after they are drawn.)
Answer:
The number of ways is [tex]\left n} \atop {}} \right. P_r = 336[/tex]
Step-by-step explanation:
From the question we are told that
The number of tickets are [tex]n = 8[/tex]
The number of finalist are [tex]r =3[/tex]
Generally the number of way by which this winners can be drawn and arrange in the order of [tex]1^{st} , \ 2nd , \ 3rd[/tex] is mathematically represented as
[tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]
substituting values
[tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]
[tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]
[tex]\left n} \atop {}} \right. P_r = 336[/tex]
