All categories

3 years ago

Find x so that m || n.

Mathematics

1 answer:

Nady [450]3 years ago

8 0

M and n are parallel
so 3x -14 = 2x +25
3x - 2x = 25 + 14
x = 39

<span>the value of x = 39 will make both these angles equal.
3(39) -14 = 117 - 14= 103
2(39) + 25 = 78 + 25 = 103

Answer x = 39</span>

You might be interested in

A distribution in which the frequency is either constantly increasing or constantly decreasing is called a(n):________

BlackZzzverrR [31]

Answer:

J Shaped

Step-by-step explanation:

5 0

3 years ago

This Quarter circle has a radius of 7 cm

Dmitry [639]

Answer:

the circumfrence is at lease 2.19 decimals

Step-by-step explanation:

3x - 14y = 92x use the circumfrence of the circle which is pi 3.14

3 0

2 years ago

Boris drove 240 miles using 9 gallons of gas. At this rate, how many gallons of gas would he need to drive 216 miles?

satela [25.4K]

Use proportion to solve this.
240 miles needs 9 gallons
216 miles needs (9÷240) × 216 = 8.1 gallons

Hope it helped!

6 0

3 years ago

Read 2 more answers

If f(x)=-x^2+2x, find f(3)-7

igomit [66]

Answer:

f(3) just means to substitute 3 in for x.

Step-by-step explanation:

f(3) just means to substitute 3 in for x.

(

)

−

Explanation:

(

)

−

(

)

According to the order of operations, first square the 3, which is 9.

(

)

−

The negative sign in front of the x squared is equivalent to multiplying by negative one.

Add -9 and 7 together, which is -2

So,

(

)

−

5 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

4 years ago