Lori wants to buy a radio for 60 dollars.
She can pay $60 now, or she can pay $12
a month for 6 months. How much more will
she pay for the radio if she makes monthly
payments?

Answer:

combined like terms and then follow  the order of operations.

Step-by-step explanation:

Answer:

D question,somewhat confusing, itsit's like simultaneous equation,but values are different

Answer:

x = 4 + 2y/3

Step-by-step explanation:

Answer:

None of the above.

1.86p + 0.7

Step-by-step explanation:

Step 1: Write expression

0.7 + p + 0.86p

Step 2: Combine like terms

0.7 + 1.86p

None of those answer choices are correct unless you wrote the problem incorrectly.

Answer:

[tex]( - \sqrt{3} \: an d \: 1)[/tex]

Answer:

6.034

Step-by-step explanation:

6 is a whole number.

.034 because it is 34 thousandths, not 34 hundredths.

Answer:

a. Pr(They both have type O)

= Pr(They both have type O)

= 0.35 x 0.45

= 0.1575 = 15.75%

b. Pr( they both have the same blood type)

= Pr( they both have the same blood type)

= 2/8

= 0.25 = 25%

c. Pr( at least one person has type O)

= Pr (at least one person has type O)

= 1 - 0.3575

= 0.6425 = 64.25%

Step-by-step explanation:

a) Data:

                   O       A         B       AB

Chinese   0.35   0.27    0.26     0.12

American 0.45   0.4      0.11       0.04

b) Calculations:

i. Pr(They both have type O)

= Probability of Chinese with O multiplied by Probability of American with O

= 0.35 * 0.45

= 0.1575 = 15.75%

ii. Pr( they both have the same blood type)

= Probability of two out of 8 outcomes

= 2/8

= 0.25 = 25%

iii. Pr( at least one person has type O)

= Probability of (1 – p(none) )

The probability of none = p(none O blood type)

= p(none)

for Chinese = (0.27 + 0.26 + 0.12) * for American ( 0.4 + 0.11 + 0.04)

= 0.65 * 0.55 = 0.3575

Pr (at least one person has type O) = 1 - 0.3575

= 0.6425

Answer:

length: 9cm

width: 9cm

Step-by-step explanation:

9×9=81

should be 2 3 andd 5

think of the - (- as a plus sign (this is what i was always taught) to add them so it would in turn be (-5) + 12 which equals 7 and choice 3 and 5 also equal this

[tex]_9C_7=\dfrac{9!}{7!2!}=\dfrac{8\cdot9}{2}=36[/tex]

Answer: -10

Step-by-step explanation: All you have to do is plug the values into the equation. -4+2(-3). Then you solve the equation using PEDMAS.

1. -4+2(-3)

2. -4+(-6)

3.-4-6

4.-10

Answer:

8

Step-by-step explanation:

-b + 2y

if

b = 4

and

y = 3

then:

-b + 2y = -4 + 2*6 = -4 + 12

= 8

Answer:

[tex] {x}^{2} + 5x + 10[/tex]

Answer:

[tex]\large \boxed{x^2 +5x+10}[/tex]

Step-by-step explanation:

A polynomial is an expression that has variables, coefficients, and constants.

An example of a polynomial can be x² - 6x + 2.

Answer:

[tex]IQR=Q_{3}-Q_{1}[/tex]

Step-by-step explanation:

The inter-quartile range is a measure of dispersion of a data set.

It is the difference between the third and the first quartile.

[tex]IQR=Q_{3}-Q_{1}[/tex]

The 1st quartile (Q₁) is well defined as the mid-value amid the minimum figure and the median of the data set. The 2nd quartile (Q₂) is the median of the data. The 3rd quartile (Q₃) is the mid-value amid the median and the maximum figure of the data set.

Answer: 1.5 less than 15 is divided by a number d.

Step-by-step explanation:

area of Arc subtending [tex]360^{\circ}[/tex] (i.e. the whole circle) is $\pi r^2$

so area of Arc subtending $\theta^{\circ}$ is, $\frac{ \pi r^2}{360^{\circ}}\times \theta^{\circ}$

$\theta =72^{\circ}$ so the area enclosed by one such arc is $\frac{\pi (10)^272}{360}$

abd there are 2 such arcs, so double the area.

[tex] \LARGE{ \underline{ \boxed{ \rm{ \purple{Solution}}}}}[/tex]

For solving this question, Let's know how to find the area of a sector in a circle?

[tex] \large{ \boxed{ \rm{area \: of \: sector = \frac{\theta}{360} \times \pi {r}^{2} }}}[/tex]

Here, Θ is the angle of sector and r is the radius of the circle. So, let's solve this question.

We have,

By using formula,

⇛ Area of shaded region = 2 × Area of each sector

⇛ Area of shaded region = 2 × Θ/360° × πr²

⇛ Area of shaded region = 2 × 72°/360° × 22/7 × 10²

⇛ Area of shaded region = 2/5 × 100 × 22/7

⇛ Area of shaded region = 40 × 22/7

⇛ Area of shaded region = 880/7 inch. sq.

⇛ Area of shaded region = 125.71 inch. sq.

☄ Your Required answer is 125.71 inch. sq(approx.)

━━━━━━━━━━━━━━━━━━━━

Answer:

Step-by-step explanation:

A probability density function (pdf) is used for continuous random variables. That is why p is between 0 and 1 (the two extremes - 0 and 1 - exclusive).

X = 100pth percentile of W

Y = 100(1-p)th percentile of W

Expressing Y as a function of X;

Y = 100(1-p)th = 100th - 100pth

Recall that 100pth is same as X, so substitute;

Y = 100th - X

where 100th = hundredth percentile of W and X = 100pth percentile of W  

Answer:

65%

Step-by-step explanation:

If 14 are male, then 26 are female.

To find the percent female, divide the number of females by the total.

26/40 = 0.65

So, the percentage of the class that is female is 65%

Answer:

C. 65%

Step-by-step explanation:

We know that of the 40 total students, 14 are male, which means the remaining students are female.

To find how many are female, we subtract 14 from 40:

40 - 14 = 26 females

Percentage is simply a part divided by a whole, multiplied by 100. Here, the "part" is the number of females, which is 26. The "whole" is the total number of students, which is 40. So, we have:

(26 / 40) * 100 = 65

The answer is thus C, 65%.

~ an aesthetics lover

Answer:

(a) 1155.84

(b) 48.2%

(c) D

Step-by-step explanation:

The number of total responses is, N = 38,528.

(a)

It is provided that 3% answered with "a lot".

Compute the actual number of responses consisting of "a lot" as follows:

n (a lot) = N × P (a lot)

            = 38528 × 0.03

            = 1155.84

Thus, the actual number of responses consisting of "a lot" is 1155.84.

(b)

The number of responses consisting of "very little or none" is,

n (very little or none) = 18,566

Compute the percentage of responses consisted of "very little or none" as follows:

[tex]P(\text{very little or none})=\frac{n(\text{very little or none})}{N}[/tex]

                                  [tex]=\frac{18566}{38528}\\\\=0.481883\\\\\approx 0.482[/tex]

The percentage is: 0.482 × 100% = 48.2%.

Thus, the percentage of responses consisted of "very little or none" is 48.2%.

(c)

As the sample size increases the sample statistic value gets closer and closer to the actual population parameter value.

Thus, making the sample statistic an unbiased estimator of the population parameter.

And proving that the sample is a true representative of the population.

Thus, the correct option is (D).

Answer:

The​ z-score for 1880 is 1.34.

The​ z-score for 1190 is -0.88.

The​ z-score for 2130 is 2.15.

The​ z-score for 1350 is -0.37.

And the z-score of 2130 is considered to be unusual.

Step-by-step explanation:

The complete question is: A standardized​ exam's scores are normally distributed. In recent​ years, the mean test score was 1464 and the standard deviation was 310. The test scores of four students selected at random are ​1880, 1190​, 2130​, and 1350. Find the​ z-scores that correspond to each value and determine whether any of the values are unusual. The​ z-score for 1880 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 1190 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 2130 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 1350 is nothing. ​(Round to two decimal places as​ needed.) Which​ values, if​ any, are​ unusual? Select the correct choice below​ and, if​ necessary, fill in the answer box within your choice. A. The unusual​ value(s) is/are nothing. ​(Use a comma to separate answers as​ needed.) B. None of the values are unusual.

We are given that the mean test score was 1464 and the standard deviation was 310.

Let X = standardized​ exam's scores

The z-score probability distribution for the normal distribution is given by;

                          Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean test score = 1464

           [tex]\sigma[/tex] = standard deviation = 310

S, X ~ Normal([tex]\mu=1464, \sigma^{2} = 310^{2}[/tex])

Now, the test scores of four students selected at random are ​1880, 1190​, 2130​, and 1350.

So, the z-score of 1880 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                                      =  [tex]\frac{1880-1464}{310}[/tex]  = 1.34

The z-score of 1190 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                                =  [tex]\frac{1190-1464}{310}[/tex]  = -0.88

The z-score of 2130 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                                =  [tex]\frac{2130-1464}{310}[/tex]  = 2.15

The z-score of 1350 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                                =  [tex]\frac{1350-1464}{310}[/tex]  = -0.37

Now, the values whose z-score is less than -1.96 or higher than 1.96 are considered to be unusual.

According to our z-scores, only the z-score of 2130 is considered to be unusual as all other z-scores lie within the range of -1.96 and 1.96.

Answer:

See below.

Step-by-step explanation:

To find roots of an equation, we use this formula:

[tex]z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})),[/tex] where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).

In this case, n = 4.

Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.

Part 2: Solving for root #1

To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by [tex]\frac{\pi}{180}[/tex] and simplify.

[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))[/tex]

[tex]z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))[/tex]

Root #1:

[tex]\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}[/tex]

Part 3: Solving for root #2

To solve for root #2, follow the same simplifying steps above but change k  to k = 1.

[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\[/tex]

Root #2:

[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}[/tex]

Part 4: Solving for root #3

To solve for root #3, follow the same simplifying steps above but change k to k = 2.

[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\[/tex]

Root #3:

[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}[/tex]

Part 4: Solving for root #4

To solve for root #4, follow the same simplifying steps above but change k to k = 3.

[tex]z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\[/tex]

[tex]z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\[/tex]

Root #4:

[tex]\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}[/tex]

The fourth roots of 16(cos 200° + i(sin 200°) are listed above.

Answer:

We conclude that no more than 10% of its microwaves need repair during the first five years of use.

Step-by-step explanation:

We are given that a maker of microwave ovens advertises that no more than 10% of its microwaves need repair during the first 5 years of use.

In a random sample of 50 microwaves that are 5 years old, 12% needed repairs.

Let p = population proportion of microwaves who need repair during the first five years of use.

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10%      {means that no more than 10% of its microwaves need repair during the first five years of use}

Alternate Hypothesis, [tex]H_A[/tex] : p > 10%     {means that more than 10% of its microwaves need repair during the first five years of use}

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 microwaves who need repair during the first 5 years of use = 12%

           n = sample of microwaves = 50

So, the test statistics =  [tex]\frac{0.12-0.10}{\sqrt{\frac{0.10(1-0.10)}{50} } }[/tex]

                                    =  0.471

The value of z-test statistics is 0.471.

Now, at a 0.04 level of significance, the z table gives a critical value of 1.751 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 0.471 < 1.751, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that no more than 10% of its microwaves need repair during the first five years of use.

Answer:

C. -8, -6, -4, -2, ...

Step-by-step explanation:

An arithmetic sequence increases by the same amount every time through addition or subtraction. There is a common difference.

A: -2, 4, -6, 8, ... If there were a common difference, the numbers would not switch between being positive and back to negative. The numbers would either keep going positive or keep going negative.

B: 2, 4, 8, 16, ... The common difference between 16 and 8 is 16 - 8 = 8. The difference between 8 and 4 is 8 - 4 = 4. Since the difference changes between the numbers, this is not an arithmetic sequence.

C. -8, -6, -4, -2, ... The common difference between -2 and -4 is -2 - (-4) = -2 + 4 = 2. The difference between -4 and -6 is -4 - (-6) = -4 + 6 = 2. The difference between -6 and -8 is -6 - (-8) = -6 + 8 = 2. Since the common difference is always two, this is an arithmetic sequence.

Hope this helps!

Answer:

4/7

Step-by-step explanation:

divide both by two for its simplest form

Answer:4/7

Step-by-step explanation

Divide both the numerator and denominator by 2

The result for the numerator is 8/2=4

that of the denominator is 14/2=7

Therefore the resultant answer is 4/7

Answer:

[tex]\huge\boxed{y = 8}[/tex]

Step-by-step explanation:

-4x + 3y = 12

Given that x = 3

-4 (3) + 3y = 12

-12 + 3y = 12

Adding 12 to both sides

3y = 12+12

3y = 24

Dividing both sides by 3

y = 8

Answer:

y =8

Step-by-step explanation:

-4x + 3y = 12

Let x = 3

-4(3) +3y = 12

-12+3y = 12

Add 12 to each side

-12+12+3y =12+12

3y =24

Divide each side by 3

3y/3 = 24/3

y =8

Answer:

7.5 cm²

Step-by-step explanation:

Dimensions of the large ∆:

[tex] base (b) = 3cm, height (h) = 9cm [/tex]

[tex] Area = 0.5*b*h = 0.5*3*9 = 13.5 cm^2 [/tex]

Dimensions of the small ∆:

[tex] base (b) = 2cm, height (h) = 6cm [/tex]

[tex] Area = 0.5*b*h = 0.5*2*6 = 6 cm^2 [/tex]

Difference between the area of the large and the small ∆ = 13.5 - 6 = 7.5 cm²

Answer:

The value is [tex]T = \$54200[/tex]

Step-by-step explanation:

From the question we are told that

      The  number of shares is  n  =  400

      The rate  of each share is  [tex]k = 135\frac{1}{2} = 135.5[/tex]

Generally the total price is mathematically represented as

     [tex]T = 400 * 135.5[/tex]

      [tex]T = \$54200[/tex]

Answer:

-11.2 pounds

Step-by-step explanation:

It is given that,

Shawna finds a study of American men that has an equation to predict weight (in pounds) from  height (in inches):

y = -210 + 5.6x

Height of Shawna's dad is 72 inches

Weight is 182 pounds

We need to find the residual of weight and height for Shawna's dad.

Predicted weight of 72 inches men,

y' = -210 + 5.6(72)

y' = 193.2 pounds

So, residual is :

Y = 182 - 193.2

Y = -11.2 pounds

So, the residual of weight and height for Shawna's dad is -11.2 pounds.

Answer:

-11.2 pounds

Step-by-step explanation:

Got it right on the test.

Answer:

60/220

Step-by-step explanation:

we use combination,

[tex] (\frac{5}{1} ) \times ( \frac{4}{1} ) \times ( \frac{3}{1} )[/tex]

[tex]5 \times 4 \times 3 = 60[/tex]

then, all divided by,

[tex] (\frac{12}{3}) = 220 [/tex]

[tex]60 \div 220[/tex]

The probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.

The probability helps us to know the chances of an event occurring.

[tex]\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]

The sample contains five milk chocolates, three dark chocolates, and four white chocolates. Therefore, the probability that the first piece is milk chocolate is

[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{5}{12}[/tex]

Now, since the chocolate is been eaten the sample size will reduce from 12 chocolates in total to 11 chocolates in total (four milk chocolates, three dark chocolates, and four white chocolates). Therefore, the probability of the second piece being white chocolate is

[tex]\rm Probability=\dfrac{\text{Number of White choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{4}{11}[/tex]

Now, as the chocolate is been eaten the sample size will reduce from 11 chocolates in total to 10 chocolates in total (four milk chocolates, three dark chocolates, and three white chocolates). Therefore, the probability of the third piece being milk chocolate is

[tex]\rm Probability=\dfrac{\text{Number of Milk choclates}}{\text{Total number of choclates}}[/tex]

[tex]\rm Probability=\dfrac{4}{10}[/tex]

Thus, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is

[tex]\rm Probability=\dfrac{5}{12}\times \dfrac{4}{11} \times \dfrac{4}{10} = \dfrac{80}{1320} = 0.06[/tex]

Hence, the probability of the first piece being milk chocolate, the second being white chocolate, and the third being milk chocolate is 0.06.

Learn more about Probability:

https://brainly.com/question/795909

Answer:

  17 by 21 inches

Step-by-step explanation:

The perimeter is twice the sum of the dimensions, and the area is their product, so you have ...

  L + W = 38

  LW = 357

__

Solution:

  W(38 -W) = 357 . . . . . substitute for L

  -(W^2 -76W) = 357 . . expand on the left

  -(W^2 -38 +19^2) = 357 -19^2 . . . . complete the square

  (W -19)^2 = 4 . . . . . . . write as a square

  W -19 = ±√4 = ±2 . . . take the square root; next, add 19

  W = 19 ±2 = {17, 21} . . . . if width is one of these, length is the other

The dimensions are 17 by 21 inches.

Answer:

Your answer is here.Enjoy dude

Answer:

12.56 unit²

Step-by-step explanation:

The form of the circle is:

Let's bring the given to the form of a circle as above:

As per the form, we got r² = 2², so the radius of circle is 2 units.

The area of circle:

Answer:

Step-by-step explanation:

Hello, by definition a perfect square can be written as [tex]a^2[/tex] where a in a positive integer.

So, to answer the first question, [tex]6^2[/tex] is a perfect square.

(a,b,c) is a Pythagorean triple means the following

[tex]a^2+b^2=c^2[/tex]

Here, it means that

[tex]x^2=20^2+21^2=841=29^2 \ \ \ so\\\\x=29[/tex]

Thank you.

Answer:

Its B

Step-by-step explanation: