A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical -value approach.= 20.5, n = 11 , σ = 7, H0: μ = 18.7; Ha: μ ≠ 18.7, α = 0.01


Answer 1


 Z = 0.8528 < 2.576

The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance

Null hypothesis is Accepted at 0.01 level of significance.

There is no significance difference between the means

Step-by-step explanation:

Given data

size of the sample 'n' = 11

mean of the sample x⁻ =20.5

Mean of the Population μ = 18.7

Standard deviation of Population σ = 7

Test statistic

                  Z = (x^(-) -mean)/((S.D)/(√(n) ) )

                  Z = (20.5 -18.7)/((7)/(√(11) ) )

                  Z = (1.8)/(2.1105)

                  Z = 0.8528

critical Value

Z_{(\alpha )/(2) } = Z_{(0.01)/(2) } = Z_(0.005) = 2.576

The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance

Null hypothesis is Accepted at 0.01 level of significance.

There is no significance difference between the means

Answer 2


B. 18.7 ± 9.7

Step-by-step explanation:

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Scores on a test in a very large class are bell-shaped and symmetric. The mean on the test was 75, and the standard deviation was 5. What percent of the scores were above 75?



50% of the scores were above 75

Step-by-step explanation:

Problems of normally distributed(bell-shaped) samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 75, \sigma = 5

What percent of the scores were above 75?

This is 1 subtracted by the pvalue of Z when X = 75. So

Z = (X - \mu)/(\sigma)

Z = (75 - 75)/(5)

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

50% of the scores were above 75

A puppy weighs 12 ounces. What fractional part of a pound is this?



The correct answer is 3/4

Because 3 × 4 = 12

Hope this helps!!!! Happy Holidays!!!! (:

The puppy is 3/4 of a pound.

Find a diginacci sequence with no equal term to 11: quick answer please



uhh hope this helps

Step-by-step explanation:

A Diginacci sequence is created as follows.

• The first two terms are any positive whole numbers.

• Each of the remaining terms is the sum of the digits of the previous

two terms.

For example, starting with 5 and 8 the Diginacci sequence is

5, 8, 13, 12, 7, 10,. . .

The calculations for this example are

5 + 8 = 13, 8 + 1 + 3 = 12, 1+ 3 +1+ 2 = 7, 1 + 2 + 7 = 10.

a) List the first 26 terms of the Diginacci sequence above.

b) Find, with explanation, two starting terms for a Diginacci sequence

so that its 2021st term is 11.

c) Find, with explanation, a Diginacci sequence that has no term equal

to 11.

d) Find, with explanation, a sequence with two different starting terms

which contains five consecutive terms that are even and not all identical

Both 2 and 3 is correct.

A submarine at - 45 feet dives 50 feet. Write an expression to represent the submarine's elevation. Need help


-150+(-45)= -195

Negative since it’s deeper

this might be yo answer im not sure

i hope this helped u tho

Difficult Math. Up to the challenge?



We have the three angles of a triangle as an equation, so we can add them together, and they equal 180.


4x+10+12x-6+3x+5= 19x+9=180

19x=171 ... so: x= 171/19 = 9.

So ABC = 102 .. Obtuse angle

BAC = 46 .. Acute angle

BCA = 32 .. Acute Angle


Angle 1 = 180 - BAC = 180 - 46 = 134

Angle 2= 180 - ABC = 180 - 102 = 78

A set X is said to be closed under multiplication if for every X1, X2 E X we have X1X2 E X. Let A be the union of all bounded subsets X CR that are closed under multiplication. Does inf(A) exist? If it does, find it.



inf(A) does not exist.

Step-by-step explanation:

As per the question:

We need to prove that A is closed under multiplication,

If for everyX_(1), X_(2)\in X

X_(1)X_(2)\in X


Suppose, x, y \in A

Since, both x and y are real numbers thus xy is also a real number.

Now, consider another set B such that:

B = {xy} has only a single element 'xy' and thus [B] is bounded.

Since, [A] represents the union of all the bounded sets, therefore,

B\subset A

⇒ xy \in A

Therefore, from x, y \]in A, we have xy \]in A.

Hence, set a is closed under multiplication.

Now, to prove whether inf(A) exist or not


Let us assume that inf(A) exist and inf(A) = \beta

Thus \beta is also a real number.

Let C be another set such that

C = { \beta - 1}

Now, we know that C is a bounded set thus { \beta - 1} is also an element of A

Also, we know:

inf(A) =  \beta


n(A)\geq \beta


\beta - 1 is an element of A and  \beta - 1 \leq \beta

This is contradictory, thus inf(A) does not exist.

Hence, proved.