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A study conducted in 2000 found that the mean number of children under 18 per household in a certain community was 1.7. A statis

Margarita [4]

Answer:

c). Two tailed test

Step-by-step explanation:

The given hypothesis are

Null hypothesis: H0:μ= 1.7

Alternative hypothesis: H1:μ≠ 1.7

The alternative hypothesis demonstrates that mean number of children are not 1.7 in 2000. This means that mean number of children can be greater than 1.7 or mean number of children can be less than 1.7. Thus, the given alternative hypothesis indicates the two tailed test.

3 0

3 years ago

Please help! If i don't get all my homework done I can't leave the house, but I'm stuck on this question.

TEA [102]

Plug y=-4x+5 into a website called desmos, it will give you a graph! The y-intercept is 5 and it’s at a slope of 4/1

7 0

3 years ago

You need to figure out how much money you make each week at the sales job. Assume you work a

Over [174]

Answer:

This true because why ?

Step-by-step explanation:

3 0

2 years ago

A glass paperweight has a composite shape: a square pyramid fitting exacty on top of an 8 centimeter cube. The pyramid has a hei

ArbitrLikvidat [17]

Answer:

Part 1) The volume of the paperweight is $576\ cm^{3}$

Part 2) The total surface area of the paperweight is $400\ cm^{2}$

Step-by-step explanation:

Part 1) what is the volume of the paperweight?

we know that

The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube

step 1

Find the volume of the pyramid

The volume of the pyramid is equal to

$V=\frac{1}{3}BH$

where

B is the area of the square base

H is the height of the pyramid

$B=8^{2}=64\ cm^{2}$

$H=3\ cm$

substitute

$V=\frac{1}{3}(64)(3)=64\ cm^{3}$

step 2

Find the volume of the cube

The volume of the cube is equal to

$V=b^{3}$

$V=8^{3}=512\ cm^{3}$

step 3

Find the volume of the paperweight

$64\ cm^{3}+512\ cm^{3}=576\ cm^{3}$

Part 2) what is the total surface area of the paperweight?

we know that

The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid

step 1

Find the surface area of 5 faces of the cube

$SA=5b^{2}$

$SA=5(8^{2})=320\ cm^{2}$

step 2

Find the lateral area of the pyramid

$LA=4[\frac{1}{2}bh]$

$LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}$

step 3

Find the total surface area of the paperweight

$320\ cm^{2}+80\ cm^{2}=400\ cm^{2}$

8 0

3 years ago

Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t

NISA [10]

Answer:

Class second have higher score and have greater spread.

Step-by-step explanation:

For first box plot

$\text{Minimum value }= 53$

$Q_1=62$

$Median=80$

$Q_3=86$

$\text{Maximum value }= 89$

$Range=Maximum-Minimum=89-53=36$

$IQR=Q_3-Q-1=86-62=24$

For second box plot

$\text{Minimum value }= 56$

$Q_1=62$

$Median=74$

$Q_3=89$

$\text{Maximum value }= 96$

$Range=Maximum-Minimum=96-56=40$

$IQR=Q_3-Q-1=89-62=27$

First class has greater minimum value, it means first class has lower grades.

First quartile of both classes are same, it means equal number students in both classes have less than 62 marks.

First class has greater median.

second class has greater third quartile.

Second class has greater Maximum value. It means second class have higher score than first class.

Second class has greater range. It means the data of second class has greater spread.

Second class has greater inter quartile range. It means the data of second class has greater spread.

Therefore, the class second have higher score and have greater spread.

5 0

3 years ago

Please help Please help please