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3 years ago

PLSSSS HELP ME!!!

1 answer:

7 0

Amount of juice hold by Cup B which is in the shape of a cylinder having width 2 inches that is radius 1 inches and height 7 inches

= cubic inches

Amount of juice hold by cup A which is in the shape of a cone having width 2 inches that is radius 1 inches and height 3 inches

= cubic inches

Amount of juice that cup B will hold than cup A when both are completely full = cubic inches

= 6 × 3.14

= 18.84 cubic inches

Option A: 18.8 cubic inches

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Exam parallelogram ABCD segments CE and BE have lengths 15-X and 2r what is the value of x

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The whole triangle is 180° , and means + in math

15-x + 2x= 180

-x+2x=180-15

-x+2x=165

2x=165

2x/2=165+/2

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A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple

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Answer:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=46$ represent the mean height for the sample

$s=5.148$ represent the sample standard deviation

$n=5$ sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:

Null hypothesis: $\mu = 40$

Alternative hypothesis: $\mu \neq 40$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing we got:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

6 0

3 years ago