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3 years ago

BRAINLIEST AND 20 POINTS

1 answer:

3 0

1. Which ratios form a proportion?
a. 4/5, 20/25 -----> 4/5, 4/5 Yes
b.8/12, 18/24 -----> 2/3, 3/4 No
c. 1/3, 7/24 ------> 8/24,7/24 No
d. 2/5, 6/16 -----> 2/5, 3/8 No

2.Which ratio forms a proportion with 9/15 = 3/5
a.3/6 = 2/3 No
b.2/3 No
c.12/30 = 2/5 No
d. 6/10 = 3/5 Yes

3.Which proportion has cross products of 5 x 24 and 8x15?
a. 5/8=15/24 ----> 5 * 24 = 8 * 15 Yes
b.5/24=8/15 ----> 5 * 15 = 8 * 24 No
c.8/24=15/5 ----> 5 * 8 = 15 * 24 No
d. 15/8=5/24 ----> 15 * 24 = 5 * 8 No

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A classroom measures 14 meters by 7 meters by 3.5 meters. How many cube meters of air are in the room?

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3 years ago

Mandy is selling pancakes and waffles at her diner. On Monday she sold a total of 300 pancakes and waffles. If she charges $3 pe

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Answer:

Mandy sold 250 pancakes and 50 waffles.

Step-by-step explanation:

From the information given, you can say that the sum of pancakes and waffles is equal to 300:

x+y=300

Also, the statement indicates that she charges $3 per pancake, $1 per waffle, and made a total of $800 and you can express this as:

3x+y=800

You have the following equations:

x+y=300 (1)

3x+y=800 (2), where:

x is the number of pancakes she sold

y is the number of waffles she sold

Now, you can solve for x in (1):

x=300-y (3)

Then, you can replace (3) in (2):

3(300-y)+y=800

900-3y+y=800

900-800=2y

100=2y

y=100/2

y=50

Finally, you can replace the value of y in (3) to find x:

x=300-50

x=250

According to this, the answer is that Mandy sold 250 pancakes and 50 waffles.

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2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

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Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

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3 years ago