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Sphinxa [80]
2 years ago
7

I got to hurry so i kinda didnt have time

Mathematics
1 answer:
k0ka [10]2 years ago
5 0

Answer:

b

Step-by-step explanation:

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Someone help with this problem and the other ones if you can!!!!
SOVA2 [1]

Answer:

sorry i cant I really need the points tho

8 0
2 years ago
Alan is connecting three garden hoses to make one longer hose.The green hose is 6.25 feel long the orange hose is 5.755 feet lon
nevsk [136]

We will use addition to find the length of the new, longer hose.

6.25 feet + 5.755 feet + 6.5 feet = 18.505 feet is the length of the new longer hose.

8 0
3 years ago
A company compiles data on a variety of issues in education. In 2004 the company reported that the national college​ freshman-to
nasty-shy [4]

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

8 0
3 years ago
In politics, marketing, etc. we often want to estimate a percentage or proportion p. One calculation in statistical polling is t
aksik [14]

The absolute value inequality is given as |(p - 0.43)I ≤ 0.019

<h3 />

<h3>How to describe the proportion using the absolute value inequality</h3>

The proportion p = 43% = 0.43

Margin of error = 1.9% = 0.019

The value of the proportion can then be said to lie between

(0.43 - 0.019) ≤ p ≤ (0.43 + 0.019)

In order to convert to the absolute inequality we would be having

-0.019 ≤ (p - 0.43) ≤ 0.019

I (p - 0.43)I ≤ 0.019

Read more on margin of error here

brainly.com/question/24289590

#SPJ1

3 0
2 years ago
JIM'S BACKYARD IS A RECTANGLE THAT IS 15 5/6 YARDS LONG AND 10 2/5 YARDS WIDE. JIM BUYS SOD IN PIECES THAT ARE 1 1/3 YARDS LONG
Stolb23 [73]
Jim's backyard:
15 5/6 yd long
10 2/5 yd wide

Sod pieced:
1 1/3 yd long
1 1/3 yd wide

Area of backyard:
15 5/6 times 10 2/5 =
164 2/3 yd

Area of Sod Piece:
1 1/3 times 1 1/3 =
1 7/9 yd

Divide area of backyard by area of sod piece :
164 2/3 divided by 1 7/9 =
92 5/8

He will need to buy 93 pieces of sod to cover his backyard.
7 0
2 years ago
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