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OLga [1]
3 years ago
15

Pls answer this question plssss

Mathematics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

241/9

Step-by-step explanation:

27=243/9

243/9-(4/2)/9=241/9

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There are 2112 people in a concert.
nadya68 [22]

Answer:

75% of 352 children are boys

Step-by-step explanation:

If 5/6 of the people participaticipants to a concert are adults then 1 out of 6 prople are children so we may reach a number of 2112:6=352 children

if 25% of the children are girls then 75% are boys

88×3=264 boys

7 0
3 years ago
Julle goes to several grocery stores and researches the price of a 12 oz bottle of juice.
lisabon 2012 [21]

Answer:

i thr 10 dollers would be ear by 1 thousand

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3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%2064%7D%20" id="TexFormula1" title=" \sqrt{ - 64} " alt=" \sqrt{ - 64} " ali
LenKa [72]

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<h3>8i</h3>

Step-by-step explanation:

4 0
2 years ago
How many trolley cars are needed to hold all the passengers? Write as a mixed number
Oksanka [162]

Answer:

46\frac{1}{8}

Step-by-step explanation:

<u><em>Formulate</em></u>

<u><em /></u>1845\div40

<u><em>Calculate</em></u>

<u><em /></u>\frac{1845}{40}

<u><em>Cross out the common factor</em></u>

<u><em /></u>\frac{369}{8}

<u><em>Write </em></u>\frac{369}{8} <u><em>as a mixed fraction</em></u>

<u><em /></u>46\frac{1}{8}

<em>I hope this helps you</em>

<em>:)</em>

7 0
3 years ago
The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi
Vesnalui [34]

Answer:

(a) X\sim N(\mu = 73, \sigma = 16)

(b) 0.7910

(c) 0.0401

(d) 0.6464

Step-by-step explanation:

Let <em>X</em> = amount of time that people spend at Grover Hot Springs.

The random variable <em>X</em> is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable <em>X</em> is:

X\sim N(\mu = 73, \sigma = 16)

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

P(X

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

P(60

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0
3 years ago
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