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3 years ago

Please help dood pls

Mathematics

1 answer:

Zinaida [17]3 years ago

5 0

Answer:

$\Large \boxed{\mathrm{\bold{B.} } \ x=7, \ y = 4\sqrt{2} }$

Step-by-step explanation:

We can split the shape into a right triangle and a rectangle.

tan θ = opp/adj

Let c be the missing base for the right triangle.

tan45 = 4/c

c = 4/tan45

c = 4

x = 3 + 4

x = 7

sin θ = opp/hyp

sin45 = 4/y

y = 4/sin45

y = $4\sqrt{2}$

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Urgentttt please❤️ Urgent

prisoha [69]

Answer: a = 110ᴼ

Step-by-step explanation:

The angles on each line of the figure add to equal 180ᴼ

180 - 70 = 110

a = 110ᴼ

b and 70ᴼ are vertical angles, so their measures are congruent.

b = 70ᴼ

180 - 70 = 110

c = 110ᴼ

3 0

3 years ago

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.