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ankoles [38]
3 years ago
10

Please hurry! Giving branniest to correct answer! 6TH GRADE MATH!

Mathematics
2 answers:
Stells [14]3 years ago
8 0
A. C. D. is the answers :)
Mnenie [13.5K]3 years ago
4 0

Answer: A C D

Step-by-step explanation:

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a survey indicated that 5 out of 8 doctors used brand X aspirin. If 4000 doctors were surveyed, how many used brand X?
fredd [130]

Answer:

2500 doctors

Step-by-step explanation:

4000/8= 500

500x5=2500

8 0
3 years ago
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
Solve for h. <br> 7/5h - 2= -3/5h +7<br> Please show work!
netineya [11]
7/5h+3/5h-2=7
10/5h-2=7
2h=11
h=5.5
3 0
3 years ago
Read 2 more answers
Which of the below is not one of the properties of logarithms?
MrRa [10]

Answer:

option-B

Step-by-step explanation:

we know that

Sum rule of logarithm:

log(a)+log(b)=log(a\times b)

which is same as

the log of a product (ab) is equal to the addition of log a nad log b

Subtraction rule of logarithm:

log(a)-log(b)=log(\frac{a}{b} )

which is same as

the log of the quotient of a and b is equal to the log of a minus the log of b

Exponent rule of logarithm:

log(a^b)=blog(a)

which is same as

the log of the quantity a raised to b is equal to the product of b and the log of a

so,

option-B is not correct

8 0
3 years ago
A community organizes a phone tree in order to alert each family of emergencies. In the first stage, one person calls five famil
masya89 [10]
Answer: It will take 7 stages for all 19,530 families to be called.

In the first stage, only 5 families are called.

In the second stage, we would have 5 x 5 = 25 families being called.

In the third stage, we would have 5 x 5 x 5 = 125 families being called.
We can use exponents to finish the list and see when our goal is reached.

4th stage: 5^4 = 625

5th stage: 5^5 = 3125

6th stage: 5^6 = 15625

7th stage: 5^7 = 78,125
6 0
3 years ago
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