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maw [93]
3 years ago
13

Helpppp plsssssssssssssss

Mathematics
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer: Section B

Step-by-step explanation: A rhombus is a parallelogram.

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What two numbers that you multiply to equal 96?
Lady_Fox [76]

3 and 32

Step-by-step explanation: 32 *3 equals 96

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3 years ago
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Use a half-angle identity to find the exact value
Tatiana [17]

Given:

\cos 15^{\circ}

To find:

The exact value of cos 15°.

Solution:

$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}

Using half-angle identity:

$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}

$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}

Using the trigonometric identity: \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}

            $=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}

Let us first solve the fraction in the numerator.

            $=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}

Using fraction rule: \frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}

            $=\sqrt{\frac {2+\sqrt{3}}{4}}

Apply radical rule: \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}

           $=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}

Using \sqrt{4} =2:

           $=\frac{\sqrt{2+\sqrt{3}}}{2}

$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}

5 0
3 years ago
(a) An angle measures 27°. What is the measure of its complement?
larisa [96]

Answer:

a) 63°

b) 127°

Step-by-step explanation:

Complements are 90° while supplements are 180°.

So you need to find the complement of 27°. So, 90-27=63

To find the supplement, it is 180-53=127

6 0
3 years ago
A pair of boots were $45 they were on sale for $38.25 what percent were they off
Mashutka [201]

Answer:

15% off

Step-by-step explanation:

4 0
3 years ago
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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
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