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2 years ago

Twenty more than the number n is 125. What is the value of n? 145 125 105 20

Mathematics

2 answers:

DENIUS [597]2 years ago

7 0

Answer:

105

Step-by-step explanation:

twenty more that the number n is 125 so n+20=125
subtract twenty on both sides
125–20=n
125–20=105
n=105

The wording is really word but I'm pretty sure this is correct

sweet-ann [11.9K]2 years ago

6 0

Answer:

105

Step-by-step explanation:

n=105 as 20 more than 105 is 125. Hope it helps!

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1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the

Musya8 [376]

Answer:

A) 34.13%

B) 15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

$z=\frac{x-m}{s}$

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

$z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0$

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

= 0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

$P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587$

C) Between 80 and 120

$P( 80$

D) less than 80

$P( x < 80 ) = P( z$

E) Between 70 and 100

$P( 70$

F) More than 130

$P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013$

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2 years ago

Please answer is question in two minutes

Kryger [21]

Answer:

C: 38 units

hope this helps!!

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3 years ago

Troy received a gift of 100 shares of stock from his grandmother on July 1 of the current year. Troy's grandmother had owned the

kumpel [21]

Troy has a basis of $39 per share

total value of shares is $3900

Step-by-step explanation:

On July 1, the date of the gift, the stock was selling for $39 per share.

Troy's basis in the stock = current value of shares

No need to consider the base rate which is in 15 years ago

The current rate of 100 shares = $39 × 100 = $3900

Troy has a basis of $39 per share

total value of shares is $3900

4 0

3 years ago

Mr. Carter drove 638 miles the first weekend. He drove 243 miles the second weekend.

zaharov [31]

638-243 = 395 miles

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7 0

2 years ago

Read 2 more answers

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o

Vikki [24]

Answer:

(a) The percent of her laps that are completed in less than 130 seconds is 55%.

(b) The fastest 3% of her laps are under 125.42 seconds.

Step-by-step explanation:

The random variable X is defined as the number of seconds for a randomly selected lap.

The random variable X is normally distributed with mean, μ = 129.71 seconds and standard deviation, σ = 2.28 seconds.

Thus, $X\sim N(129.71,\ 2.28^{2})$ .

(a)

Compute the probability that a lap is completes in less than 130 seconds as follows:

$P(X$

$=P(Z$

The percentage is, 0.55 × 100 = 55%.

Thus, the percent of her laps that are completed in less than 130 seconds is 55%.

(b)

Let x represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of z for the above probability is:

z = -1.88

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42$

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let x₁ and x₂ be the values between which the middle 80% of the distribution lie.

That is,

$P(x_{1}$

The value of z for the above probability is:

z = 1.28

Compute the values of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-1.28=\frac{x_{1}-129.71}{2.28}\\x_{1}=129.71-(1.28\times 2.28)\\x=126.7916\\x\approx 126.80$ $z=\frac{x_{2}-\mu}{\sigma}\\1.28=\frac{x_{2}-129.71}{2.28}\\x_{2}=129.71+(1.28\times 2.28)\\x=132.6284\\x\approx 132.63$

Thus, the middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

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3 years ago