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pogonyaev
2 years ago
7

The heights of adult females are normally distributed. If you were to construct a histogram of 40 randomly selected​ women, what

shape would the histogram of those heights have and what pattern would you expect in a normal quantile plot of these​ data?
a. The histogram would by approximately bell-shaped, and the normal quantile plot would have data points that would be bell-shaped.
b. The histogram would by non-symmetric, and the normal quantile plot would have data points that would be non- linear.
c. The histogram would by approximately bell-shaped, and the normal quantile plot would have data points that would be non-linear.
d. The histogram would be approximately bell-shaped, and the normal quantile plot would have data points have follow a straight-line pattern.
Mathematics
1 answer:
Anna11 [10]2 years ago
7 0

Answer:

d. The histogram would be approximately bell-shaped, and the normal quantile plot would have data points have follow a straight-line pattern.

Step-by-step explanation:

Since the variable is normally distributed, the histogram of women's height should be approximately bell shaped (if the data was obtained form a random sample).

Again, the variable is normally distributed, therefore, the quantile plot should follow a straight line pattern (a diagonal line to be more precise).

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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
MakcuM [25]

Answer:

A=8.4063u^{2}

Step-by-step explanation:

Be the functions:

y=\frac{3}{x};y=\frac{3}{x^{2}}:x=7

according the graph:

\int\limits^1_7 {\frac{3}{x} } \, dx -\int\limits^1_7 {\frac{3}{x^{2} } } \, dx =3\int\limits^1_7 {\frac{1}{x} } \, dx -3\int\limits^1_7 {\frac{1}{x^{2} } } \, dx=3(\int\limits^1_7 {\frac{1}{x} } \, dx -\int\limits^1_7 {\frac{1}{x^{2} } } \, dx)=3[lnx-\frac{1}{x}](1-7)=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}

6 0
2 years ago
A side of a regular six - sided polygon is 8cm long. The perimeter of a similar polygon is 60cm. What is the length of a side of
Lerok [7]

Answer:

  10 cm

Step-by-step explanation:

The larger polygon is also a regular 6-sided polygon, so each side is 1/6 of the perimeter length:

  (60 cm)/6 = 10 cm . . . length of a side

4 0
2 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
2 years ago
You have just received an inheritance of $28,000 and would like to invest it into an account. The bank offers two investment pla
grigory [225]
First we need to calculate annual withdrawal of each investment
The formula of the present value of an annuity ordinary is
Pv=pmt [(1-(1+r)^(-n))÷(r)]
Pv present value 28000
PMT annual withdrawal. ?
R interest rate
N time in years
Solve the formula for PMT
PMT=pv÷[(1-(1+r)^(-n))÷(r)]

Now solve for the first investment
PMT=28,000÷((1−(1+0.058)^(−4))
÷(0.058))=8,043.59
The return of this investment is
8,043.59×4years=32,174.36

Solve for the second investment
PMT=28,000÷((1−(1+0.07083)^(
−3))÷(0.07083))=10,685.63
The return of this investment is
10,685.63×3years=32,056.89

So from the return of the first investment and the second investment as you can see the first offer is the yield the highest return with the amount of 32,174.36

Answer d

Hope it helps!
6 0
2 years ago
Ariana and Emily are both standing in line at Papa Joe’s Pizza. Ariana orders 4 large cheese pizzas and 1 order of breadsticks.
Eva8 [605]

The cost of 1 large pizza is $7.99 and cost of one breadstick is $2.50

Elimination method was used.

Step-by-step explanation:

Let,

Cost of one large pizza = x

Cost of one breadstick = y

According to given statement;

4x+y=34.46    Eqn 1

2x+y=18.48     Eqn 2

Subtracting Eqn 2 from Eqn 1

(4x+y)-(2x+y)=34.46-18.48\\4x+y-2x-y=15.98\\2x=15.98

Dividing both sides by 2

\frac{2x}{2}=\frac{15.98}\\x=7.99

Putting x=7.99 in Eqn 2

4(7.99)+y=34.46\\31.96+y=34.46\\y=34.46-31.96\\y=2.50

The cost of 1 large pizza is $7.99 and cost of one breadstick is $2.50

Elimination method was used.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/2131336
  • brainly.com/question/2150928

#LearnwithBrainly

6 0
2 years ago
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