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jeka94
2 years ago
12

The test score for a class are shown. What is the average test score? 79,80,92,92,81,100,88,98,71,100,91,90

Mathematics
1 answer:
Oksanka [162]2 years ago
5 0

Answer:

The average score is equal to 88.5

Step-by-step explanation:

we know that

An average test score is the sum of all the scores on an assessment divided by the number of test-takers.

so

The sum is equal to

Sum=[79+80+92+92+81+100+88+98+71+100+91+90]=1,062

The number of test-takers is n=12

The average score is equal to

\frac{S}{n}

substitute

\frac{1,062}{12}=88.5

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Answer:

See Below

Step-by-step explanation:

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3 years ago
Molly can drive her car 112 miles on 4 gallons of gas and 182 miles on 6.5 gallons write an equation to show the relationship be
Rashid [163]

Answer:

Molly can drive her car 112 miles on 4 gallons of gas and 182 miles on 6.5 gallons write an equation to show the relationship between the number of gallons she has and the number of miles she can drive and how many gallons she would need to drive 420 miles⊂⇄∪↓Ф

Step-by-step explanation:

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3 years ago
It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de
scoray [572]

Answer:

a) P(X

P(z

b) P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

P(z>-0.583)=1-P(Z

c) P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) z=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

X \sim N(72000,12000)  

Where \mu=72000 and \sigma=12000

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability using excel or the normal standard table and we got:

P(z

Part b

P(X>65000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

And we can find this probability using the complement rule and excel or the normal standard table and we got:

P(z>-0.583)=1-P(Z

Part c

P(X>100000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

And we can find this probability using the complement rule and excel or the normal standard table and we got:

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.1   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

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3 years ago
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Katena32 [7]

Answer:

Δ JKL is similar to Δ ABC ⇒ D

Step-by-step explanation:

Similar triangles have equal angles in measures

In ΔABC

∵ m∠A = 15°

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∴ m∠A + m∠B + m∠C = 180°

→ Substitute the measures of ∠A and ∠B

∵ 15 + 120 + m∠C = 180

→ Add the like terms in the left side

∴ 135 + m∠C = 180

→ Subtract 135 from both sides

∴ 135 - 135 + m∠C = 180 - 135

∴ m∠C = 45°

The similar Δ to ΔABC must have the same measures of angles

If triangles ABC and JKL are similar, then

m∠A must equal m∠J

m∠B must equal m∠K

m∠C must equal m∠L

∵ m∠J = 15°

∴ m∠A = m∠J

∵ m∠L = 45°

∴ m∠C = m∠L

∵ m∠J + m∠K + m∠L = 180°

→ Substitute the measures of ∠J and ∠L

∵ 15 + m∠K + 45 = 180

→ Add the like terms in the left side

∴ 60 + m∠K = 180

→ Subtract 60 from both sides

∴ 60 - 60 + m∠K = 180 - 60

∴ m∠K = 120°

∴ m∠B = m∠K

∴ Δ JKL is similar to Δ ABC

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3 years ago
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Arisa [49]

Answer:

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Step-by-step explanation:

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3 years ago
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