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Alika [10]
3 years ago
13

The football team manager brought a jug of water to practice. The jug holds 1,280 ounces of water. The players' water bottles ea

ch hold
15 ounces. How many water bottles can be filled from the 1,280-ounce jug?

A)80
B)85
C)86
D)92
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
8 0
I’m not sure but I think it can fill 85 bottles. Because you divide 1280 by 15 and you get 85.33 which is rounded to 85.
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You want to know the side length of the square swimming pool. Explain how you can use the perimeter then find the side length to
olganol [36]

Its a square pool and its perimeter is 100 yards.

We know that a square has 4 sides with the same length so we can give a square perimeter by the sum of their 4 sides

Lets call the side "l" and since we have 4 of them we need to sum it 4 times

l + l + l + l = perimeter

4l = perimeter

We know the perimeter

4l = 100

Divide both sides by 4

4l/4 = 100/4

l = 25

So the side length is equal to 25 yards

4 0
3 years ago
True or false: A triangle can be made by segments that are 15.9 cm, 10 cm and 25.9 cm.
Gala2k [10]

Answer:

false

Step-by-step explanation:25.9cm is too big to connect to 15.9cm and 10 cm

4 0
2 years ago
Solve for z N (17+ x ) = 34x - r
BigorU [14]

Answer:

Step-by-step explanation:

N(17+x)=34x−r

17n+xn=34x-r

xn= 34x - r - 17n

xn-34x= -r - 17n

x(n-34)= -r - 17n

x= (-r - 17n)/(n-34)

7 0
3 years ago
TRUE OR FALSE!!! PICTURE IS SHOWN
schepotkina [342]
False

distance = square root [(x2−x1)^2+(y2−y1)^2]

5 0
3 years ago
An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.
Harlamova29_29 [7]

Answer:

(A) 0.04

(B) 0.25

(C) 0.40

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)=  \frac{10}{20}=\frac{1}{2}\\P(W)=  \frac{2}{20}=\frac{1}{10}

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

6 0
3 years ago
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