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4 years ago

Can someone plz help me solved this problem I need help ASAP plz help me! Will mark you as brainiest!

Mathematics

1 answer:

lakkis [162]4 years ago

8 0

Answer:

2 & 8

-4 & -4

Step-by-step explanation:

x and y are the numbers, as per question:

x= 4+2y and xy= 16

y(4+2y)= 16
2y²+4y=16
y²+2y - 8= 0
y²+2y+1 =9
(y+1)²=9
y+1=3 ⇒ y= 2 ⇒ x= 4+2*2= 8
y+1= -3 ⇒ y= -4 ⇒ x= 4+2*(-4)= -4

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) e number of viewers ordering a particular pay-per-view program is normally distributed. Past history shows that 33.00% of the

lubasha [3.4K]

Answer:

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

Step-by-step explanation:

Problems of normally distributed samples are 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.

Past history shows that 33.00% of the time fewer than 20,000 people order the program

This means that X = 20000 has a pvalue of 0.33. So when X = 20000, Z = -0.44.

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

$-0.44 = \frac{20000 - \mu}{\sigma}$

$20000 - \mu = -0.44\sigma$

$\mu = 20000 + 0.44\sigma$

Only ten percent of the time do more than 28,000 people order the program.

This means that X = 28000 has a pvalue of 1-0.1 = 0.9. So when X = 28000, Z = 1.28.

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

$1.28 = \frac{28000 - \mu}{\sigma}$

$28000 - \mu = 1.28\sigma$

$\mu = 28000 - 1.28\sigma$

Also

$\mu = 20000 + 0.44\sigma$

$20000 + 0.44\sigma = 28000 - 1.28\sigma$

$1.72\sigma = 8000$

$\sigma = \frac{8000}{1.72}$

$\sigma = 4651.16$

$\mu = 20000 + 0.44\sigma = 20000 + 0.44*4651.16 = 22046.5$

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

6 0

3 years ago

Whats the answer how do you simplify this

Margarita [4]

A^(-b) = 1 / (a^b), so

5^(-3) = 1 / (5^3) = 1 / 125 = 0.008

7 0

3 years ago

Is this correct I need help if it's not thank you

Korolek [52]

Yes it is i stared at it for a while AND DID THIS

7 0

4 years ago

Read 2 more answers

The average watermelon weighs 8 lbs with a standard deviation of 1.5. Find the probability that a watermelon will weigh between

Andrew [12]

Answer:

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = 0.5932

Step-by-step explanation:

Step 1:-

by using normal distribution find the areas of given x₁ and x₂

Given The average watermelon weighs 8 lbs

μ = 8

standard deviation σ = 1.5

I) when x₁ = 6.8lbs and μ = 8 and σ = 1.5

$z_{1} = \frac{x_{1} -mean}{S.D} = \frac{6.8-8}{1.5} = - 0.8$

ii) when x₂ = 9.3 lbs and μ = 8 and σ = 1.5

$z_{2} = \frac{x_{2} -mean}{S.D} = \frac{9.3-8}{1.5} = 0.866>0$

<u>Step2</u>:-

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = A(z₂) - A(-z₁)

= A(0.866) - A(-0.8)

= A(0.866)+ A(0.8)

check below normal table

= 0.3051 + 0.2881

= 0.5932

<u>Conclusion</u>:-

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = 0.5932

7 0

3 years ago

Slader "Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose that we win $2 for ea

kolezko [41]

Answer:

the probabilities associated with each possible value for X

for x = -2

0.308

for x = -1

0.176

for x=0

0.011

for x =+1

0.3516

for x=+2

0.088

Step-by-step explanation

See this attachment

4 0

3 years ago