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VladimirAG [237]
3 years ago
12

Suppose y varies directly as x, and y = 21 when x = 3. Find x when y = 42.

Mathematics
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

x = 24

Step-by-step explanation:

y₂ - y₁

42 - 21 = 21

x + 21

3 + 21 = 24

So when y increases by 21, so does x.

x = 24

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BRAINLIEST, help, thank youu!!
mihalych1998 [28]

Answer: B - 15/40

Step-by-step explanation: You can simplify the answer from

3/8 x 5 = 15/40

Hope this helps ty if you do give brainliest. means alot.

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2 years ago
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Which subsets of real numbers does -22 belong
deff fn [24]
Negative number because -22
3 0
3 years ago
Please help me: Ten athletes ran two races of the same length. The scatter plot shows their times. Identify each of these statem
Lemur [1.5K]

The scatter plot has been attached

Answer:

Options C, D & E are true

Step-by-step explanation:

Option A is wrong because from the scatter plot, only four athletes were faster in the second race than in the first one.

Option B is wrong because only 1 athlete had his second race time differing from the first race time by exactly 2 seconds.

Option C is true because exactly 9 of the times for the first race were at least 16 seconds

Option D is true because there are exactly 3 athletes who had the same time in both races

Option E is true because 8 of the times for the second race were less than 17 seconds

4 0
2 years ago
Variables x and y are in direct proportion, and y = 12.5 when x = 25. If x = 40, then y =
BabaBlast [244]
25/12.5 = 40/y
cross multiply
(25)(y) = (12.5)(40)
25y = 500
y = 500/25
y = 20 <====
6 0
3 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
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