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3 years ago

PQR is a right-angled triangle. PQ = 16 cm. PR = 8 cm. Calculate the length of

Mathematics

1 answer:

vlada-n [284]3 years ago

8 0

SOLUTION:

PQR is a right-angle triangle.

Therefore, to solve this problem, we will use Pythagoras theorem which is only applicable to right-angle triangles.

Pythagoras theorem is as displayed below:

a^2 + b^2 = c^2

Where c = hypotenuse of right-angle triangle

Where a and b = other two sides of right-angle triangle

Now we will simply substitute the values from the problem into Pythagoras theorem in order to obtain the length of QR.

c = PQ = 16cm

a = PR = 8cm

b = QR = ?

a^2 + b^2 = c^2

( 8 )^2 + b^2 = ( 16 )^2

64 + b^2 = 256

b^2 = 256 - 64

b^2 = 192

b = square root of ( 192 )

b = 13.8564...

b = 13.86 ( to 2 decimal places )

FINAL ANSWER:

Therefore, the length of QR is 13.86 centimetres to 2 decimal places.

Hope this helps! :)
Have a lovely day! <3

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3 years ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

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