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3 years ago

-40+2n=4n-8n+8 EXPLAIN IN NUMBER FORM AND NO WORDS

Mathematics

1 answer:

marin [14]3 years ago

8 0

Answer:

−40+2n=−4n+8

−40+2n+40=−4n+8+40

2n=−4n+48

2n+4n=−4n+48+4n

6n=48

answer: n=8

Step-by-step explanation:

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The box and whiskers plot shows the weights, in kilograms, of randomly selected packages in a mail truck. Which of the following

GREYUIT [131]

Answer:2

Step-by-step explanation:

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3 years ago

CPM homework 8.1.2 problem 8-19

Kisachek [45]

The area of the figure is 45 sq.mi and the perimeter is 23.5354 mi.

Area is the space occupied by a two dimensional object,

The area and perimeter of the given shape can be determined by dividing it into basic shapes.

The area and the perimeter of the shape given = Area of the rectangle +Area of upper triangle - Area of two small triangles at the bottom

Area of rectangle = 5 mi * 9 mi = 45 sq.mi

Area of upper Triangle = (1/2) *9*3 = 27/2 sq.mi

Area of small Triangle = (1/2) * 3*6 = 9 sq.mi

Area of other small Triangle = (1/2)*3*3 = 9/2 sq.mi

Area of the figure = 45 +13.5 -9-4.5

Area of the figure = 45 sq.mi

Similarly for Perimeter = Perimeter of the rectangle +Perimeter of upper triangle -Perimeter of two small triangles at the bottom

Perimeter = 28 -(3+6+6.708) -(3+3+4.2426) +(3+9+9.486

Perimeter = 23.5354 mi

Therefore , The area of the figure is 45 sq.mi and the perimeter is 23.5354 mi.

To know more about Area

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2 years ago

Round up 395,621 to the nearest ten thousand

jeyben [28]

396,000
because 5 goes up and anything lower than five stays the same

4 0

3 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

choli [55]

Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) $n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation

n=2341 represent the random sample taken

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.55$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}}$

Part a) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523$

$0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577$

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$

And rounded up we have that n=1831

8 0

4 years ago

What is the approximate volume of the composite figure? Use 22/7 for pi. Round to the nearest hundredth.

Temka [501]

Answer:

188.57 cm³

Step-by-step explanation:

Half sphere = 1/2 volume sphere

Sphere = 4/3 * 22/7 * 3³

Sphere = 4/3 * 22/7 * 27/1

Sphere = 2376/21 = 113.14

113.14 / 2 = 56.57

Cone = 22/7 * 3² * h/3

Cone = 22/7 * 9/1 * 14/3

Cone = 2772/21 = 132

132 + 56.57 = 188.57 cm³

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4 0

3 years ago