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stepan [7]
2 years ago
13

5. A quarter is tossed in the air 300 times.

Mathematics
2 answers:
tankabanditka [31]2 years ago
3 0

Answer:

C

Step-by-step explanation:

The probability of getting heads or tails is 50%.

300*50% = 150

guapka [62]2 years ago
3 0

Answer:

C. 150 times

Step-by-step explanation:

Because a coin only has tails and heads, you can say that the probability would be half. Because it was tossed 300 times, you can say: 300*0.5=150

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Anybody know the answers to this frl it’s due in 15 mins
hammer [34]

Answer:

4. j(11) = 8

I think it is linear

5. g(-3) = 18

No, it's a parabola since x is to the second power

6. f(-7) = 5

No, I think it's an absolute value function

7. h(0) = 13

I think it is linear, just the slope is 0

Step-by-step explanation:

3 0
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
2 years ago
What is the value of this expression?
Mila [183]

Answer:

1/4

Step-by-step explanation:

1)2²= 4

2)4²= 16

3)3x2/3=2

4)6+2-4/8=1/4

4 0
2 years ago
⎧
d1i1m1o1n [39]

<u>Given</u>:

The given expression to find the nth term of the sequence is d(n)=d(n-1) \cdot (-5)

The first term of the sequence is d(1)=8

We need to determine the third term of the sequence.

<u>Second term:</u>

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

d(2)=d(2-1) \cdot (-5)

d(2)=d(1) \cdot (-5)

d(2)=8 \cdot (-5)

d(2)=-40

Thus, the second term of the sequence is -40.

<u>Third term:</u>

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

d(3)=d(3-1) \cdot (-5)

d(3)=-40 \cdot (-5)

d(3)=120

Thus, the third term of the sequence is 120.

7 0
3 years ago
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Need help 8th grade math
Yanka [14]
The answer is the 3rd one
6 0
3 years ago
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