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pantera1 [17]
3 years ago
8

#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

so

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

so

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

therefore

m\angle 2=68^o

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... 18x^2 +30x^6 = 6x^2(3 +5x^4)

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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    

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df=n-1=10-1=9  

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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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3 years ago
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