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viva [34]
3 years ago
5

How many pints are in 7.5 quarts?

Mathematics
2 answers:
nikitadnepr [17]3 years ago
7 0

Answer:

Step-by-step explanation: 69 quarts

Anna71 [15]3 years ago
3 0

Answer:

15

Step-by-step explanation:

1 quart = 2 pints

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2765 divided by 9 partial quoents
storchak [24]

Answer:307.2222 or just 307

Step-by-step explanation:

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Mrs. Lee bought 2 boxes of plastic bags. She had 1,988 craft sticks. She divided the Craft Sticks equally into 28 plastic bags f
Serjik [45]
The answer is D. 72 sticks.
5 0
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1/2 of the people who live in the neighborhood vollunteer at the comunity garden. 1/4 of the vollunteers are teenagers. What fra
ki77a [65]

Answer:

3/4

Step-by-step explanation:

With regards to the above, since 1/4 of the volunteers are teenagers, the fraction of the volunteers would be

= 1 - 1/4

= 3/4

It therefore means that 3/4 of the volunteers are teenagers.

3 0
3 years ago
A number line is divided into fifths. How many equal segments are there between the point at zero and the point 4/5?
ella [17]

Answer:

3?

Step-by-step explanation:

A little confused by the question but is it just

1/5 2/5 3/5 4/5

1     2     3     4

so theres 3 in between

7 0
3 years ago
work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
Yuki888 [10]

Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the​ conference is 0.10.

(c) The probability that Clarice attends the​ conference is 0.40.

(d) The probability that John stays​ home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (<em>M</em>), Roberto (<em>R</em>), John (<em>J</em>), Dominique (<em>D</em>) and Clarice (<em>C</em>) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

{5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

The probability of the event <em>E</em> is:

P(E)=\frac{n(E)}{N}

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of  Roberto and John is, <em>RJ</em>.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the​ conference as follows:

P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10

Thus, the probability that Roberto and John attend the​ conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the​ conference as follows:

P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40

Thus, the probability that Clarice attends the​ conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays​ home as follows:

P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60

Thus, the probability that John stays​ home is 0.60.

4 0
3 years ago
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