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2 years ago

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Four student are each trying to raise the same amount of money for a class trip. The table below shows how much of each student'

s goal has been met.

1 answer:

postnew [5]2 years ago

3 0

Answer:

the answer is 5/8, 65%, 2/3, 0.7

Step-by-step explanation:

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Nancy leaves the school and drives toward her town at 27 mph. greg leaves the same school 2 hours later and drives the same rout

Paul [167]

Nancy's rate=27 mph
Greg's rate=45 mph
t=Greg's time
t+2=Nancy's time
Since they both traveled same distance then:
27(t+2)=45t
27t+54=45t
18t=54
t=3 hours
Greg will pass Nancy after 3 hours

6 0

2 years ago

What is the sum of adding two opposites together? *

BartSMP [9]

Answer:

I'm guessing the opposites are negative and positive so whenever you add two opposite numbers it will always be 0. The answer will always be 0.

For example:

-3 + 3 = 0

4 - 4 = 0

-15 + 15

5 0

2 years ago

Use the distributive property to write an equivalent expression.<br> 2(8v+2)

andreyandreev [35.5K]

Answer:

16v+4

Step-by-step explanation:

2(8v+2) = 16v+4

4 0

1 year ago

When rolling a number cube 600 Times, predict the relative frequency of rolling a 3 or 6

Flauer [41]

Answer:

The frequency of rolling a 3 or a 6 would be the same even if you try thousands or millions times.

The probability of rolling a 3 is: 1/ total outcomes probability = 1/6

The probability of rolling a 6 is: 1/ total outcomes probability = 1/6

The probability of rolling a 6 or a 3 = 2/ total outcomes probability = 2/6 = 1/3

Therefore, if you rolling the cube 600 times, the probability of rolling a 3 or a 6 is still 1/6. If 3 and 6 are both allowed, the probability would become 1/3

Hope this helped :3

7 0

2 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = <u>0.11</u>

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

2 years ago