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3 years ago

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Corinne is 6 years younger than her brother. Which of the following expressions shows the sum of their ages if b represents Cori

nne's brother's age?A. b-6
B. 2b-6
C.2(b-6)
D. b divided by 6

1 answer:

andrey2020 [161]3 years ago

6 0

Corinne : b - 6 years old;
her brother : b years old.
The sum: b - 6 + b = 2 b - 6
Answer: B )

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Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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3 years ago

The mean of 10 values is 19. If further 5 values are

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Answer:

22

Step-by-step explanation:

Pretend the 10 values in the first sentence are a,b,c,d,e,f,g,h,i,j

Pretend the addition 5 values is k,l,m,n,o

So the mean of all the 15 data is (a+b+c+d+e+f+g+h+i+j+k+l+m+n+o)/15=20

So the sum of all 15 data is a+b+c+d+e+f+g+h+i+j+k+l+m+n+o=300 since 15(20)=300

Now let's look at the first 10: We have their mean so we can write:

(a+b+c+d+e+f+g+h+i+j)/10=19

so a+b+c+d+e+f+g+h+i+j=190 since 10(19)=190

So that means using our first sum equation and our equation sum equation we have

190+k+l+m+n+o=300

k+l+m+n+o=300-190

k+l+m+n+o= 110

So the average of those 5 numbers mentioned in your problem is 110/5=22

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3 years ago

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Slope is -4
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3 years ago

We have a null hypothesis that u=3.75, but we are reluctant to accept this as the population value in case the true value is dif

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Answer:

option A

Step-by-step explanation:

To Test

H₀ : μ = 3.75

H₁ : μ ≠ 3.75

now,

If μ = 3.95 then it means H₀ is not true.

then the probability is rejecting the null hypothesis.

Hence, the probability is the power of the test against the alternative μ=3.95

The correct answer is option A

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Answer:

8 rows and 3 extra

Step-by-step explanation:

40/5=8

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