pls help with all the questions

Answer: The pelican dove 5ft

Step-by-step explanation:

Pelican = 3ft

Fish = -2ft

Went down 3ft to get to 0, went down 2 more feet to get the fish.

Answer:

When variables are the same, multiplying them together compresses them into a single factor (variable). ... When multiplying variables, you multiply the coefficients and variables as usual. If the bases are the same, you can multiply the bases by merely adding their exponents.

Step-by-step explanation:

Answer:

C  -3 < x ≤ 4

Step-by-step explanation:

-9 < 4x + 3 ≤ 19.

Subtract 3 from all sides

-9-3 < 4x + 3-3 ≤ 19-3

-12 < 4x  ≤ 16

Divide by 4

-12/4 < 4x/4 ≤ 16/4

-3 < x ≤ 4

Answer:

395.52

Step-by-step explanation:

247.2/2.5=98.88(1 bib)

98.88x4=395.52(4 bibs)

9514 1404 393

Answer:

  dy/dx = 2x +1

Step-by-step explanation:

The power rule can be used.

  d/dx(x^n) = n·x^(n-1)

Then the derivative of x² is 2x¹ = 2x, and the derivative of x = 1x⁰ = 1.

The derivative of the function is ...

  dy/dx = 2x +1

The selected statistical questions are as follows:

A what is the typical amount of rainfall for the month of June in the Galapagos islands

D how many songs does the class usually listen to each day

G how long does it typically take for 2nd graders to walk a lap around the track

Statistical questions are known by these two main characteristics:

This means that the data vary.  They are not the same data.  For example, "How old is Monday?" is not a statistical question because it has one answer.

Thus, statistical questions are not answered by a single number or answer, especially when the correct answer is just one.  

Learn more examples of statistical questions here: https://brainly.com/question/15729334

Answer:

6.8 feet

Answer From Gauth Math

Width = area/length = 22.6/5.2 = 113 ≈/26

Diagonal = √[length² + width²] ≈ √23.0 feet

Answer:

C

Step-by-step explanation:

Domain of the function is the whole R

Answer:

picture not clear what is the length

Answer:

P = 24

Step-by-step explanation:

Since all the sides are the same length, the shape is a square.

Multiply all sides by 6.

6 cm x 4 sides = 24

Recall that

[tex]\dfrac{dv(t)}{dt} = a(t) \Rightarrow dv(t) = a(t)dt[/tex]

Integrating this expression, we get

[tex]\displaystyle v(t) = \int a(t)dt = \int(t^2 - 4t + 5)dt[/tex]

[tex]\:\:\:\:\:\:\:= \frac{1}{3}t^3 - 2t^2 + 5t + C_1[/tex]

Also, recall that

[tex]\dfrac{ds(t)}{dt} = v(t)[/tex] or

[tex]\displaystyle s(t) = \int v(t)dt = \int (\frac{1}{3}t^3 - 2t^2 + 5t + C_1)dt[/tex]

[tex]\:\:\:\:\:\:\:= \frac{1}{12}t^4 - \frac{2}{3}t^3 + \frac{5}{2}t^2 + C_1t + C_2[/tex]

Next step is to find [tex]C_1\:\text{and}\:C_2[/tex]. We know that at t = 0, s = 0, which gives us [tex]C_2 = 0[/tex]. At t = 1, s = 20, which gives us

[tex]s(1) = \frac{1}{12}(1)^4 - \frac{2}{3}(1)^3 + \frac{5}{2}(1)^2 + C_1(1)[/tex]

[tex]= \frac{1}{12} - \frac{2}{3} + \frac{5}{2} + C_1 = \frac{23}{12} + C_1 = 20[/tex]

or

[tex]C_1 = \dfrac{217}{12}[/tex]

Therefore, s(t) can be written as

[tex]s(t) = \frac{1}{12}t^4 - \frac{2}{3}t^3 + \frac{5}{2}t^2 + \frac{217}{12}t[/tex]

Answer:

D. 6.5

Step-by-step explanation:

The diameter of the cylinder is 1.25 x 2 = 2.5

SK = √1.25² + 6² = √42.25 = 6.5

Answer:

I am not so sure but I am assuming-5 2/3 to be mixed fraction so improper fraction wud be -17/3

now 4*- 17/3 = 68/3

now divide -5/8 ÷ 12

so u must know the rule that if division we can multiply by reciprocal

so -5/(8*12) = -5/96

hope that helps

9514 1404 393

Answer:

Step-by-step explanation:

Let x represent the enrollment the first year. Then x(1 -8%) = 0.92x represents the enrollment the second year. The total for the two years is ...

  x + 0.92x = 13,876

  x = 13,876/1.92 = 7227.083 ≈ 7227 . . . . students the first year

  13876 -7227 = 6649 . . . . students the second year

Answer:

The range is the set of first coordinates

Answer:

Hello,

Step-by-step explanation:

do I remind you of the formula :

where (a,b) is the vertex and y=k the directrix

[tex]y=\dfrac{(x-a)^2}{2(b-k)} +\dfrac{b+k}{2} \\\\\\y=-\dfrac{(x+4)^2}{4} -1 \\\\Using\ identification:\\a=-4\\2(b-k)=-4\\b+k=-2\\\\\left\{\begin{array}{ccc}b-k&=&-2\\b+k&=&-2\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}2b&=&-4\\2k&=&0\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}b&=&-2\\k&=&0\\\end{array}\right.\\[/tex]

Focus=(-4,-2)

Directrix: y=0

Vertex=(-4,-1)

Answer:

a) The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b) Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c) 37 cars would have to be sampled.

Step-by-step explanation:

Question a:

We have the sample standard deviation, and thus, the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.1448

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.1448\frac{6.2}{\sqrt{15}} = 3.4[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26.7 - 3.4 = 23.3 mpg.

The upper end of the interval is the sample mean added to M. So it is 26.7 + 3.4 = 30.1 mpg.

The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b. What would happen to the interval if you increased the confidence level from 95% to 99%? Explain

Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c. The lead engineer is not happy with the interval you constructed and would like to keep the width of the whole interval to be less than 4 mpg wide. How many cars would you have to sample to create the interval the engineer is requesting?

Width is twice the margin of error, so a margin of error of 2 would be need. To solve this, we have to consider the population standard deviation as [tex]\sigma = 6.2[/tex], and then use the z-distribution.

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

How many cars would you have to sample to create the interval the engineer is requesting?

This is n for which M = 2. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = 1.96\frac{6.2}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 1.96*6.2[/tex]

[tex]\sqrt{n} = \frac{1.96*6.2}{2}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*6.2}{2})^2[/tex]

[tex]n = 36.9[/tex]

Rounding up:

37 cars would have to be sampled.

Answer: 0.155

Step-by-step explanation:

g / 1000 = kilograms

155 / 1000 = 0.155

Answer:

B

Step-by-step explanation:

Answer:

Step-by-step explanation:

D={ 6  , -9  , 1 }

R={  0  ,-3 }

Answer:

Mr. Shim would record the number +10 or 10 for Ben because of the word "more".

For Gail, Mr. Shin would record the number -8 because of the word, "less''.

Since Nathaniel did not improve or decrease the number of crunches, Mr. Shin would record the number 0.

I hope this helps better

Answer:

A is the answer! I think you know because the formula is given at the top.

This is a question that asks about the advantages of a systematic random sampling. Thus, we first take a look at the types of sampling, and then we see the advantage of systematic random sampling.

Samples may be classified as:

Convenient: Sample drawn from a conveniently available pool.

Random: Basically, put all the options into a hat and drawn some of them.

Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.

Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.

Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.

Systematic:

One of the bigger advantages is that the systematic sampling eliminate clusters, which means that the last option is wrong.

Inadvertently missing patterns is a problem in systematic sampling, and not an advantage, thus the third option is also wrong.

It also does not reduce sampling variability, thus the first option is wrong.

From this, it can be concluded that the correct option is:

Systematic random sampling does not require a finite population size.

For another example of systematic random sampling, you can check https://brainly.com/question/21100042

[tex]\boxed{\sf a_n=\dfrac{(-1)^n}{5n}}[/tex]

a:-

[tex]\\ \sf\longmapsto a_1=\dfrac{(-1)^1}{5(1)}[/tex]

[tex]\\ \sf\longmapsto a_1=\dfrac{-1}{5}[/tex]

[tex]\\ \sf\longmapsto a_1=-5[/tex]

b:-

[tex]\\ \sf\longmapsto a_4=\dfrac{(-1)^4}{5(4)}[/tex]

[tex]\\ \sf\longmapsto a_4=\dfrac{1}{20}[/tex]

c:-

[tex]\\ \sf\longmapsto a_{30}=\dfrac{(-1)^{30}}{5(30)}[/tex]

[tex]\\ \sf\longmapsto a_{30}=\dfrac{1}{150}[/tex]

d:-

[tex]\\ \sf\longmapsto a_{19}=\dfrac{(-1)^{19}}{5(19)}[/tex]

[tex]\\ \sf\longmapsto a_{19}=\dfrac{-1}{95}[/tex]

Answer:

X=0,y=-5

x=4,y=-1

Step-by-step explanation:

Replace all occurrences of y with x^2-3x-5

(x^2-3x-5)+5=x

x^2-3x=x

X^2-4x=0

so :x=0,4

enter the value of x in the equation then find y

y=-5,-1

For engineering, it would be very useful to know Pythagorean theorem. You can use it to measure the tension in each ropes.