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3 years ago

#3 Can someone please help me with these questions!!!!

Mathematics

1 answer:

rosijanka [135]3 years ago

7 0

Answer:

Option A 90,68

$m\angle 1=90^o$

$m\angle 2=68^o$

Step-by-step explanation:

see the attached figure to better understand the problem

The figure shows a kite

we know that

A kite properties include

1) two pairs of consecutive, congruent sides

2) congruent non-vertex angles

3) perpendicular diagonals

Find the measure of angle 1

we have that

$m\angle 1=90^o$ ----> by the diagonals are perpendicular

Find the measure of angle 2

we know that

triangle AEB is congruent with triangle CEB

$m\angle EAB=m\angle ECB=m\angle 2$

In the right triangle EAB

$m\angle EAB+22^o=90^o$ ----> by complementary angles in a right triangle

$m\angle EAB=90^o-22^o=68^o$

therefore

$m\angle 2=68^o$

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<em>Comment on the question</em>

Since this answer is not among the answer choices, I suggest you ask your teacher to demonstrate how this problem is worked.

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Answer:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

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

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

$\bar X=59.4$ represent the sample mean

$s=3.239$ represent the sample standard deviation

$n=10$ sample size

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:

Null hypothesis: $\mu \leq 55$

Alternative hypothesis: $\mu > 55$

Replacing the info given we got:

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

And replacing the info given we got:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

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