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mina [271]
4 years ago
7

Cathy needs to collect linear data which set of data should cathy collect

Mathematics
1 answer:
Readme [11.4K]4 years ago
3 0

Option A is correct.

Cathy needs to collect linear data . She should collect the data of option A-The height of a building for 20 years.

Explanation:

Linear data is that data whose slope is always same. Here lets assume, initially the building had 2 floors. Each year 2 floors are added above the previous floors. So our coordinate will be (0,2) (1,4) (2,6) (3,8)  and so on.

So slope = \frac{difference in y}{difference in x}

= \frac{4-2}{1-0} = \frac{2}{1}

= \frac{8-6}{3-2} = \frac{2}{1}

So, here slope is 2 for all coordinates.

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Okay this is simple once you get use to it. What you first need to do is figure out the formula. I don’t have a calculator on me so I will just tell you the formula so you can get the answer
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3 years ago
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Factorise:p/4-q2/16 pls <br><br> help me with it
Annette [7]

Answer:

1/4 ( p - 1/4 q^2).

Step-by-step explanation:

p/4 - q2/16

The greatest common factor is 1/4 so we have the answer:

1/4 ( p - 1/4 q^2).

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3 years ago
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Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
3 years ago
Can someone help me figure this out
Setler79 [48]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

It is told that, 330 students that are surveyed, are 9th graders.

Hence, 0.55 represents 330 students, that is 1 represents \frac{330}{0.55} = 600 students.

The number of 9th grade students whose favorite elective is music is represented by 0.18.

1 ≡ 600

0.18 ≡600\times0.18 = 108.

The percentage of students whose favorite elective is home economics and are also in 10th grade is \frac{0.06}{1} \times100 = 6%.

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Yes!! the line AED is cut with a perpendicular line. The rule for perpendicular lines is that the angles made =90.

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