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artcher [175]
3 years ago
12

When making a statistical inference about the mean of normally distributed population based on the samples drawn from the popula

tion which of the following statements is correct all else being equal
Eight. The 68% confidence interval is wider than the 90% confidence interval.
B.The 90% confidence interval is narrower than the 95% confidence interval.
C. The 90% confidence interval is wider than the 99% confidence interval.
D.The 99% confidence interval is narrower than the 68% confidence interval.
Mathematics
1 answer:
Tresset [83]3 years ago
7 0
<span>When making a statistical inference about the mean of normally distributed population based on the samples drawn from the population which of the following statements is correct all else being equal is that The 90% confidence interval is wider than the 99% confidence interval. . The answer is letter C.</span>
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The table shows a linear function.
Elza [17]

Answer:

a) I used the equation x=-5; f(x)=-11 and x=-4; f(x)=-3. We get the range = -3-(-11)=8

b) I used the equation x=-5; f(x)=-11 and x=-3; f(x)=5. We get the range = 5-(-11)=16

c) I used the equation x=-5; f(x)=-11 and x=-2; f(x)=13. We get the range = 13-(-11)=24

d) Range of input is equal with the ratios of the output. You can find the pattern of output above 8,16,and 24 can divided by 8 and give the result 1,2 and 3

4 0
3 years ago
Right answers only!!!
Nata [24]

Step-by-step explanation:

7

8 0
3 years ago
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The repairs bill on your car includes $355 for parts $146 for labor and a sales tax of $40 what is the total amount you owed
marissa [1.9K]

Answer:

$541

Step-by-step explanation:

Given:

The repair bill includes $355 for parts $146 for labor and a sales tax of $40

Question asked :

what is the total amount we owed = ?

Solution:

Total cost of parts to repair = $355

Total cost of labor = $146

Sales tax =$40

By adding all the cost:

Total cost of repair = $355 + $146 + $40

                                = $541

Therefore, total amount we have to pay = $541

7 0
3 years ago
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gavmur [86]

Answer:

the base is 10 and the height is 6.4

3 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean \mu =266\\

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find P(X\geq 308)

Using normal distribution,

z=\frac{X-\mu}{\sigma}

z=\frac{308-266}{15}

=\frac{42}{15}

Thus z=2.8

So P\left (X\geq 308  \right )=1-P(X

=1-P(z

=1-Table\:  value\:  of\:  2.8

=1-0.99744

=0.00256

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu

This implies X=-1.8808*15+266=237.788

Thus the babies who are born on or before 238 days are considered to be premature.

5 0
3 years ago
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