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4 years ago

What is the interquartile range of the data set? {48, 50, 65, 60, 80, 42, 45, 50, 42, 55, 45}

Mathematics

1 answer:

Tcecarenko [31]4 years ago

6 0

The answer would ether be 16.5 or 10. Sorry, I'm not sure haven't done interquartile range in a little while.

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Please Help? A circle has an arc length of 5π in. The central angle for this arc measures π/3 radians. What is the area of the a

Zielflug [23.3K]

Answer:

The area of the associated sector is $\frac{25}{24}\pi \ in^{2}$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The circumference of a circle is equal to

$C=2\pi r$

we have

$C=5\pi\ in$

substitute and solve for r

$5\pi=2\pi r$

$r=2.5\ in$

step 2

Find the area of the circle

we know that

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=2.5\ in$

substitute

$A=\pi (2.5^{2})=6.25\pi\ in^{2}$

step 3

Find the area of the associated sector

we know that

$2\pi\ radians$ subtends the complete circle of area $6.25\pi\ in^{2}$

by proportion

Find the area of a sector with a central angle of $\pi/3\ radians$

$\frac{6.25\pi }{2\pi} =\frac{x}{\pi/3}\\x=6.25*(\pi/3)/2\\ \\x=\frac{25}{24}\pi \ in^{2}$

8 0

3 years ago

How many grams must be adde to 970g to make 1 and a half kg

madam [21]

Answer:

530 grams

Step-by-step explanation:

970 grams = 0.970 kg

1 kg = 1000 grams

Thus: 500+30 = 530 grams

3 0

3 years ago

Last one i think if not i will post the last one this is confusing

maria [59]

Answer: 23

Step-by-step explanation: Because you do 34-11

5 0

3 years ago

Read 2 more answers

Pls need help 100 points and crown if right!!!

nekit [7.7K]

Answer:

$\sf 2\dfrac{1}{4}\ cups \ of \ sugar$

Explanation:

Let the full batch be x

Here given:

3/4 cup of sugar required to make 1/3 batch of cookies

Build equation:

$\sf \rightarrow \dfrac{1}{3}x = \dfrac{3}{4} \ cup \ of \ sugar$

Solve:

$\sf \rightarrow x = \dfrac{3(3)}{1(4)}$

$\sf \rightarrow x = \dfrac{9}{4}$

$\rightarrow \sf x = 2\dfrac{1}{4}$

5 0

1 year ago

Read 2 more answers

On a particular production line,the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What

KengaRu [80]

Answer:

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

Step-by-step explanation:

Consider the provided information,

Let X is the number of defective bulbs.

Ten light bulbs are randomly selected.

The likelihood that a light bulb is defective is 5%.

Therefore sample size is = n = 10

Probability of a defective bulb = p = 0.05.

Therefore, q = 1 - p = 1 - 0.05 = 0.95

Mean of binomial random variable: $\mu=np$

Therefore, $\mu=10(0.05)=0.5$

Variance of binomial random variable: $\sigma^2=npq$

Therefore, $\sigma^2=10(0.05)(0.95)=0.475$

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

5 0

3 years ago