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2 years ago

A triangle PQR is right angled at R, with PQ=80cm and R=60cm. find QR? pythagorus theorem

Mathematics

1 answer:

algol132 years ago

5 0

Answer:

Solution given:

A triangle PQR is right angled at R, with hypotenuse{h}PQ=80cm

and

base[b]PR=60cm.

perpendicular [P]= QR

by using Pythagoras law

h²=p²+b²

80²=QR²+60²

QR²=80²-60²

QR= $\sqrt{2800}$

QR=20 $\sqrt{7}$ =52.9=53cm

QR=53cm.

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2 years ago

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

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Answer:

a) $\mu_{\bar x} =\mu = 276.1$

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$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

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Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

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2 years ago

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3 years ago

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