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4 years ago

15

Sue has 18 CDs and 30 DVDs. He wants to place them onto the greatest number of shelves so that each shelf has the same number of

CDs and the same number of DVDs. How many shelves will he use?

1 answer:

frez [133]4 years ago

8 0

Cds 3 dvds 5

Sue will use 6 shelves with 3 cds on each and 5 dvds on each.

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The internal measures of a cuboidal room are 9m, 5m and 6m respectively. Find the total cost of

kumpel [21]

Answer:

<h3>₹ 1290</h3>

Step-by-step explanation:

The total surface area of the room = 2(LW + WH + LH)

TSA = 2(9*5 + 9*6 + 5*6)

TSA = 2(45 + 54 + 30)

TSA = 2(129)

TSA = 258

Hence the total surface area of the room is 258m²

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The cost of white washing if the ceiling of the room is also whitewashed is ₹ 1290

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3 years ago

a. Harry started a new job where he earns the same amount every hour. Last week, he worked 9 hours and earned $108. Write the li

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Answer:

I=12

Step-by-step explanation:

y=108

I=108/9

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3 years ago

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Fencing cost $10 per foot. Frances had a rectangular garden that is 4 feet wide and 5 feet long . How much money will frances ne

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Answer: $180

4ft + 5ft + 4ft + 5ft = 18ft
18 x 10 = 180

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3 years ago

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What are the solutions of the quadratic equation (x + 3)(x +3) = 49? Ax = -2 and x = -16 Bx = 2 and x = -10 Cx = 4 and x = -10 D

Ivahew [28]

Answer:

4 and -10

Step-by-step explanation:

$\displaystyle (x + 3)(x +3) = 49 \\ x^2+3x+3x+9=49 \\ x^2 +6x+9=49 \\ x^2 + 6x + 9 - 49 = 0 \\x^2+6x-40=0 \\\\ \Delta=b^2-4ac \\ \Delta=6^2-4 \cdot 1 \cdot (-40) \\ \Delta=36+160 \\ \Delta=196 \\ \\ X_{1,2}=\frac{-b \pm \sqrt{\Delta} }{2a} \\ \\ X_1=\frac{-b+\sqrt{\Delta} }{2a} = \frac{-6+14}{2} = \frac{8}{2}=4 \\ \\ X_2=\frac{-b-\sqrt{\Delta} }{2a} = \frac{-6-14}{2}=\frac{-20}{2} = -10$

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3 years ago

In a random sample of n1 = 156 male Statistics students, there are x1 = 81 underclassmen. In a random sample of n2 = 320 female

Scilla [17]

Answer:

The p-value for the test of hypothesis is P(z<-3.617)=0.0002.

Step-by-step explanation:

Hypothesis test on the difference between proportions.

The claim is that the percent of males who are underclassmen stats students (π1) is less than the percent of females who are underclassmen stats students (π2).

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2$

The male sample has a size n1=156. The sample proportion is p1=81/156=0.52.

The female sample has a size n2=221. The sample proportion in this case is p2=221/320=0.69.

The weigthed average of proportions p, needed to calculate the standard error, is:

$p=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}=\dfrac{81+221}{156+320}=\dfrac{302}{476}= 0.63$

The standard error for the difference in proportions is:

$\sigma_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.63*0.37}{156}+\dfrac{0.63*0.37}{320}}\\\\\\\sigma_{p1-p2}=\sqrt{\dfrac{0.2331}{156}+\dfrac{0.2331}{320}}=\sqrt{0.001503871+0.000728438}=\sqrt{0.002232308}\\\\\\\sigma_{p1-p2}=0.047$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_1-p_2}{\sigma_{p1-p2}}=\dfrac{0.52-0.69}{0.047}=\dfrac{-0.17}{0.047}=-3.617$

The P-value for this left tailed test is:

$P-value = P(z$

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3 years ago

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