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cupoosta [38]
2 years ago
15

Five friends—Allison, Beth, Carol, Diane, and Evelyn— have identical calculators and are studying for a statistics exam. They se

t their calculators down in a pile before taking a study break and then pick them up in random order when they return from the break. What is the probability that at least one of the five gets her own calculator?
Mathematics
1 answer:
brilliants [131]2 years ago
7 0

Answer:

20%

Step-by-step explanation:

this is because that 1 in 5 is equal to 20%

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A city has 3 new houses for every 9 old houses. If there are 21 new houses in the city, how many old houses are there?
kipiarov [429]
To find your answer you would divide 21 by 3 which would be 7, once you've got 7 you would multiply it by 9 which would give you the amount of old houses that there is which would be 63.
5 0
3 years ago
Slips of paper numbered 1 to 15 are placed in a box. A slip of paper is drawn at random. What is the probability that the number
monitta

Given:

Slips of paper numbered 1 to 15 are placed in a box.

To find:

The probability that the number picked is either a multiple of 5 or an odd number.

Solution:

We have,

Total outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

No. of total outcomes = 15

Multiple of 5 are 5, 10, 15.

Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15.

Number that are either a multiple of 5 or an odd number are 1, 3, 5, 7, 9, 10, 11, 13, 15.

No. of favorable outcomes = 9

We know that,

\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

\text{Probability}=\dfrac{9}{15}

\text{Probability}=\dfrac{3}{5}

\text{Probability}=0.6

Therefore, the  probability that the number picked is either a multiple of 5 or an odd number is 0.6.

8 0
3 years ago
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
2 years ago
The freezer at The Beech Café must be kept below -19 °C. It is currently at -17.5 °C. Is it cold enough?
Fofino [41]
It is not cold enough

3 0
3 years ago
Read 2 more answers
True or false An invalid statistical test measure may have questions that are misinterpreted.
Norma-Jean [14]
True. An invalid statistical test measure may have questions that are misinterpreted.
7 0
3 years ago
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