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xeze [42]
3 years ago
14

Solve for "n"! 3(n+4)=-6+15n help ASAP RANDOM ANSWERS WILL BE MODERATED!

Mathematics
1 answer:
amm18123 years ago
4 0
We are told to solve for n given the equation 3(n + 4) = -6 + 15n. We can solve for n by first using the distributive property, which gives us 3n + 12 = -6 + 15n. Now, subtract 15n from both sides, giving us -12n + 12 = -6. Next, subtract 12 from both sides to get -12n = -18. Now, divide both sides by -12 to get n = 18/12, or 3/2 when simplified. Therefore, the value of n in the equation 3(n + 4) = -6 + 15n is 3/2. Hope this helped!
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On his outward journey, Ali travelled at a speed of s km/h for 2.5 hours. On his return journey, he increased his speed by 4 km/
den301095 [7]

Answer:

3.08 km/h.

Step-by-step explanation:

We know that,

Speed=\dfrac{Distance}{Time}

Ali traveled at a speed of s km/h for 2.5 hours.

Let d be the distance and t be the time.

2.5=\dfrac{d}{t}

2.5t=d      ...(1)

On his return journey, he increased his speed by 4 km/h and saved 15 minutes. So, distance is d and times is t-15.

4=\dfrac{d}{t-15}

4(t-15)=d      ...(2)

From (1) and (2), we get

4(t-15)=2.5t

4t-60-2.5t=0

1.5t=60

t=40

Put t=40 in (1).

d=2.5(40)=100

So, t=40 and d=100.

Now,

Total distance = d + d = 100 + 100 = 200

Total time = t + t - 15 = 40 + 40 - 15 = 65

So, the average speed is

\text{Average speed}=\dfrac{\text{Total distance}}{\text{Total time}}

\text{Average speed}=\dfrac{200}{65}

\text{Average speed}\approx 3.08

Therefore, the average speed is 3.08 km/h.

3 0
3 years ago
Simplify (x − 4)(3x2 − 6x + 2). 3x3 + 6x2 − 22x + 8 3x3 − 18x2 + 26x − 8 3x3 − 18x2 − 22x − 8 3x3 + 6x2 + 22x + 8
kogti [31]

Answer:

3x^3-18x^2+26x-8

Step-by-step explanation:

(x-4)(3x^2-6x+2)

3x^3-6x^2+2x-12x^2+24x-8

3x^3-18x^2+26x-8

5 0
3 years ago
Solve the attached question<br><br>#yohaniJaman<br><br><br><br>​
aksik [14]

HETY is a parallelogram.

HT and EY are diagonals. We know that diagonals divides the parallelogram into two equal parts.

So ar(HET) = ar(HTY)

And, ar(HEY) = ar(EYT) now, in AHET, diagonal EY bisects the line segment HT and also the AHET,

∴ar(AHOE) = ar(AEOT)

Similarly in AETY

ar(ΔΕΟΤ) = ar(ΔΤΟΥ)

And in AHTY,

ar(ATOY) = ar(AHOY)

That means diagonals in parallelogram divides it into four equal parts.

Hence Proofed.

5 0
2 years ago
4/-7 + -6/5 <br><br>In simplest form​
nlexa [21]

See attachment for math work and answer.

6 0
3 years ago
An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish
otez555 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!

8 0
3 years ago
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