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2 years ago

This is a question I have on my homework. Cynthia favorite clothing store is having a 30% off sale. What fraction represents 30%

Mathematics

2 answers:

Illusion [34]2 years ago

3 0

The answer is 3/10...

OverLord2011 [107]2 years ago

3 0

3/10 is the answer to that question

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Find the nth term of: 11, 18, 37, 74, 135

storchak [24]

The nth term is 6n^2-11n+16

6 0

1 year ago

Write 12,430,090in expanded form

stira [4]

10,000,000+2,000,000+
400,000+30,000+90

7 0

1 year ago

What are the solution(s) to the quadratic equation 40 − x2 = 0?

Anni [7]

X = 20
First, you move the 40 to the other side.
-2x=-40
Then, you divided -40 by -2.
x= -40/-2
x= 20
x is 20 because the negatives cancel out making it positive.

4 0

1 year ago

Read 2 more answers

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

Andru [333]

Answer:

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Step-by-step explanation:

Values (x) ∑(Xi-X)^2

----------------------------------

29.1 0.1702

28.5 1.0252

28.8 0.5077

29.4 0.0127

29.8 0.0827

30.1 0.3452

30.6 1.1827

----------------------------------------

236.1 3.4088

Mean = 236.1 / 8 = 29.51

$S_{x}=\sqrt{3.4088/(8-1)}=0.6978$

Statement of the null hypothesis:

H0: u ≥ 30 the mean wedding cost is not less than $30,000

H1: u < 30 the mean wedding cost is less than $30,000

Test Statistic:

$t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861$

Test criteria:

SIgnificance level = 0.05

Degrees of freedom = df = n - 1 = 8 - 1 = 7

Reject null hypothesis (H0) if

$t$

Finding in the t distribution table α=0.05 with df=7, we have

$t_{0.05,7}=2.365$

$t>-t_{0.05,7}$ = -1.9861 > -2.365

Result: Fail to reject null hypothesis

Conclusion: Do no reject the null hypothesis

u ≥ 30 the mean wedding cost is not less than $30,000

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Hope this helps!

5 0

2 years ago

Ben swims 50,000 yards per week in his practices. Given this amount of training, he will swim the 100-yard butterfly in 51.5

Elanso [62]

Answer:

MArginal productivity: $\frac{dt}{dL}=-0.0002$

We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

Step-by-step explanation:

The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.

In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.

The marginal productivity can be expressed as:

$\frac{dt}{dL}$

where dt is the variation in time and dL is the variation in training yards.

We can not derive the function because it is not defined, but we can approximate with the last two points given:

$\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002$

Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.

6 0

1 year ago