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3 years ago

Try this please if you can. Thanks!

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

<h3>Answer:</h3>

PR ≈ 14.50

<h3>Step-by-step explanation:</h3>

Since ∠Q is a right angle, the area can be found from ...

... A = (1/2)bh = (1/2)(QP)(PR)

From trigonometry, we know ...

... QP = PR·cos(P) = PR·cos(36°)

... QR - PR·sin(P) = PR·sin(36°)

Substituting these values and the given area into the area formula gives ...

... 50 = (1/2)(PR·cos(36°))(PR·sin(36°))

... 100/(sin(36°)cos(36°)) = (PR)²

... PR = 10/√(sin(36°)cos(36°))

... PR ≈ 14.5015

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Quotient is to divide.

Let n be the unknown number.

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2 years ago

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3 years ago

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3 years ago

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insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of

ss7ja [257]

Answer: $(13.22\%,\ 19.34\%)$

Step-by-step explanation:

Given : Sample space : n= 559

Sample proportion : $\hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628$

$=16.28\%$

Significance level : $\alpha= 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Confidence level for population proportion:-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)$

Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= $(13.22\%,\ 19.34\%)$

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3 years ago