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Slav-nsk [51]
2 years ago
8

Please help with my homework

Mathematics
2 answers:
nlexa [21]2 years ago
8 0

The user @Hrishii is correct about the value of angle B.

Once you know all three angles, you can use the law of sines to find the missing sides.

----------------------------------------------

Let's use the law of sines to find side b

b/sin(B) = a/sin(A)

b/sin(22) = 1/sin(18)

b = sin(22)*1/sin(18)

b = 1.21225240111352

b = 1.21  rounding to two decimal places

----------------------------------------------

Repeat for side c

c/sin(C) = a/sin(A)

c/sin(140) = 1/sin(18)

c = sin(140)*1/sin(18)

c = 2.08010440004024

c = 2.08  rounding to two decimal places

==============================================

The fully solved triangle is

A = 18 degrees

<h3>B = 22 degrees</h3>

C = 140 degrees

a = 1

<h3>b = 1.21</h3><h3>c = 2.08</h3>

decimal values are approximate

Lina20 [59]2 years ago
3 0

Step-by-step explanation:

\angle \: A  +  \angle \: B  +   \angle \:  C  = 180 \degree \\  \therefore \:  18 \degree  +  \angle \: B  +  140 \degree = 180 \degree  \\ \therefore \:  \angle \: B  +  158 \degree = 180 \degree  \\ \therefore \:  \angle \: B  = 180 \degree  - 158 \degree  \\   \huge \red{ \boxed{\therefore \:  \angle \: B  = 22 \degree}}

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3 years ago
If RS = 17, ST = x + 6, and RT = 3x - 5, What is ST.
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3 0
3 years ago
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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