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Shtirlitz [24]
3 years ago
10

Turn these equations into Standard Form. Pls show the steps on how you did it.

Mathematics
2 answers:
Katarina [22]3 years ago
6 0

let me help you step by step

pantera1 [17]3 years ago
6 0
1 y+2x=-3/2
2y+4=-3
4x+2y=-3

2. y-4/5x=-10
5y-4x=-50
-4x+5y=-50
4x-5y=50

3. y-11/8x=6
8y-11x=48
-11x+8y=48
11x-8y=-48

4. y-1/3x=-2
3y-x=-6
-x+3y=6

5. y+6/5x=-3
5y+6x=-15
6x+5y=-15
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WRITE 7/10 AS A PERCENTAGE
IRINA_888 [86]
70% hope this helps :)
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3 years ago
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Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The p
Ymorist [56]

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

Z = \frac{X - \mu}{s}

X = 7.1

Z = \frac{7.1 - 6.3}{0.3}

Z = 2.67

Z = 2.67 has a pvalue of 0.9962

X = 5.5

Z = \frac{5.5 - 6.3}{0.3}

Z = -2.67

Z = -2.67 has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

Z = \frac{6.7 - 6.3}{0.3}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

X = 5.9

Z = \frac{5.9 - 6.3}{0.3}

Z = -1.33

Z = -1.33 has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

5 0
3 years ago
WILL GIVE BRAINLIEST PLEASE ANSWER If a circle is dilated by 5/3 and the area of the larger circle is 100pi square cm, then what
Simora [160]

Answer:

We have a small circle, that is dilated by a factor of 5/3 creating a larger circle.

If the smaller circle has a diameter D, then after the dilation, the larger circle will have a diameter equal to: (5/3)*D

We know that the area of the larger circle is:

A = 100*pi cm^2.

And the area of a circle of diameter d, is:

a = pi*(d/2)^2

knowing that the diameter of the large circle is (5/3)*D, we can find the value of D.

A = pi*( (5/3)*D/2)^2 = 100*pi cm^2

let's solve this for D:

pi*( (5/3)*D/2)^2 = 100*pi cm^2

( (5/3)*D/2)^2 = 100 cm^2

( (5/3)*D/2) = √(100 cm^2) = 10cm

D/2 = (3/5)*10cm

D = 2*(3/5)*10cm = 12cm.

Then the area of the smaller circle will be:

a = pi*(D/2)^2 = pi*(12cm/2)^2 = pi*(6cm)^2 = pi*36 cm^2

and pi = 3.14

a = pi*36 cm^2 = 3.14*36cm^2 = 113.04 cm^2

7 0
3 years ago
Diana saves $5.25 cach week. How<br> much will she save in a year? There<br> are 52 weeks in a year.
dmitriy555 [2]

go on a calculator and multiply 5.25 by 52

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The product of 4 and the depth of the pool
finlep [7]

Answer:

product means multiplication (×) so what ever is the depth of the pool is multiplied by 4

Step-by-step explanation:

what is the depth of the pool?

8 0
3 years ago
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