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1 year ago

Charmaine pays $848 for landscaping rocks for a project. The cost per pound is 32 cents. How many pounds did she purchase?

2 answers:

6 0

Okay so since it is 32 cents, it will be 0.32
You need to find the number of pounds
So, do 848/0.32
You get 2650
Charmaine purchased 2,650 pounds of rocks
hope this helps:)

DerKrebs [107]1 year ago

6 0

Answer:

answer would be d

Step-by-step explanation:

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Sheleah and her family are planning a trip from Los Angeles, California to Melbourne, Australia. While booking her family's plan

Shkiper50 [21]

Answer:

Step-by-step explanation:

According to the graph, the amount of gasoline after 16 hours is 0 gallons. ⇒ Jet has to land to refuel after 15 hours of flying.

3 0

1 year ago

What is the best estimate for 7.8 x 37.86?<br> A. 46<br> B. 50<br> C. 300<br> D. 400

lana66690 [7]

I think it’s C since it’s 295.308

8 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

Gretchen made $56,750 last year. She paid $1,200 in student loan interest and made a $3,000 contribution to her IRA. On her fede

tia_tia [17]

d. Adjustments

Studen loan interests and IRA contributions are deductions found under the heading of ADJUSTMENTS TO INCOME to compute for the Adjusted Gross Income or AGI.

Standard deductions are those based on the filing status of the individual and not his total itemized deductions. Regardless of the actual expenses incurred by an individual, he can claim a standar deduction if he is single, head of household, married filing separately, married filing jointly, qualifying widow(er). at the time he files for his federal tax return.

taxable income is the income left from all the necessary deductions.

For example: Gretchen's income => $56,750
   less: Adjustments to income
   student loan interest $1,200
   IRA Contribution 3,000 - 4,200
   ===========
Taxable income $52,550

8 0

1 year ago

Read 2 more answers

The following table shows the number of hours some teachers in two schools expect students to spend on homework each week: Schoo

Lerok [7]

Answer:

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Step-by-step explanation:

Part A

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set in four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

Interquartile range of the data is

$IQR=Q_3-Q_1=16-6.5=9.5$

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

Interquartile range of the data is

$IQR=Q_3-Q_1=15.5-8=7.5$

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.

4 0

2 years ago