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9966 [12]
3 years ago
12

PLEASE HELP!

Mathematics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

24.7619048 %

Step-by-step explanation:

The percent spent on food

food/total

520/2100

.247619048

This is in decimal form

Multiply by 100 to change it to percent

24.7619048 %

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(3a - 15) - (9a + 2) \\  - 6a - 17

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Step-by-step explanation:

Put the values of x = 6, y = -7 and z = 0.8 to the expressions:

a) 5x → 5(6) = 30

b) 3y → 3(-7) = -21

c) 10z → 10(0.8) = 8

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Help me on this math question
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3 years ago
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The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth
Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = P(Laptop) = p_{X} = \frac{4}{9}

The probability of selecting a desktop is = P(Desktop) = p_{Y} = \frac{5}{9}

Then both X and Y follows Binomial distribution.

X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})

The probability function of a binomial distribution is:

P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: {n\choose k}=\frac{n!}{k!(n-k)!}

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\  =0.304832\\\approx0.305

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

P(Y\geq 3)=1-P(Y

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0
3 years ago
A local business has an area reserved behind the store for a parking lot that is 78 meters long by 19 meters wide. The stalls of
S_A_V [24]

The total area available for cars to park will be 858m².

<h3>How to calculate the area?</h3>

The area of the parking lot will be:

= 78 × 19

= 1482m²

The area of aisle will be:

= 8 × 78

= 624m²

The available area will be:

= 1482m² - 624m²

= 858m².

The compact parking spaces that will fit in the lot will be:

N × 12.5 = 858

N = 858/12.5

N = 68

The non compact parking spaces that will fit in the lot will be:

N × 16.5 = 858

N = 858/16.5

N = 52

Learn more about area on:

brainly.com/question/7438648

#SPJ1

5 0
2 years ago
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