If m∠2 = 137 and m∠P = 22, what is m∠O?

If you want to study which science course students liked the most during high school, you should survey... science teacher's, be

den301095 [7]

Answer:

the last option

Step-by-step explanation:

please put brainliest

5 0

2 years ago

Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean $\mu=510$ ,

Standard deviation $\sigma=100$

Sample mean $x=300$

Percentage of students scored below 300 is given by,

$P(Z\leq \frac{x-\mu}{\sigma})\times 100$

$=P(Z\leq \frac{300-510}{100})\times 100$

$=P(Z\leq \frac{-210}{100})\times 100$

$=P(Z\leq-2.1)\times 100$

$=0.0179\times 100$

$=1.79\%$

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

$P(x\leq t)=0.90$

Now, $P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90$

$P(Z< \frac{t-\mu}{\sigma})=0.90$

$\frac{t-\mu}{\sigma}=1.28$

$\frac{t-510}{100}=1.28$

$t-510=128$

$t=128+510$

$t=638$

Score puts someone in the 90th percentile is 638.

5 0

4 years ago

Number 3

sdas [7]

Answer: Choice A) 1990 to 1991

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Explanation:

Let's go through the answer choices one by one.

For choice A, we go from the year 1990 to 1991 which is a change of 1 year. The wage goes from 3.80 to 4.25 over this timespan, which is an increase of 4.25-3.80 = 0.45 dollars. Divide the change in wage (0.45 dollars) over the change in change in time (1 year) to get 0.45/1 = 0.45; this indicates that the wage increased by 0.45 dollars per year over the timespan 1990 to 1991

Now onto choice B. We have a wage increase of 3.80-3.35 = 0.45 over a course of 9 years (since 1990-1981 = 9) so 0.45/9 = 0.05 is the rate of wage growth, meaning that the wage bumps up by a nickel each year. So far choice A is the winner.

Moving onto choice C, we have a wage increase of 0.25 dollars (3.35-3.10 = 0.25) over an 1 year period (1981-1980 = 1) so the rate of change for this slice of time is 0.25/1 = 0.25 dollars per year. Choice A is still the winner.

Finally, for choice D, this is over a year as well (1980-1979 = 1) and the wage increases by 0.20 dollars (3.10-2.90 = 0.20) leading to a rate of change to be 0.20/1 = 0.20

So choice A has the largest rate of change which is the same as saying it has the largest rate of wage increase. This shows why choice A is the answer.

3 0

3 years ago

Which of the following is equivalent to the expression below for x≠6?

nata0808 [166]

Answer:

The fourth answer choice is the correct one: 2(x - 3)

Step-by-step explanation:

2x²-18x+36/x-6 should be factored, as follows:

2(x²-9x+18) / (x-6) = 2(x - 3) (x - 6) / (x - 6)

Substituting x = 6 would result in division by zero and is thus not allowed.

Remembering this, we can reduce 2(x - 3) (x - 6) / (x - 6) to 2(x - 3) for x≠6.

6 0

3 years ago

The age of a father is 2 less than 7 times the age of his son. In 3 years, the sum of their ages will be 52. If the son’s presen

s2008m [1.1K]

Answer:

First,

the age of the son: s

the age of the father is 2 less than 7 times the age of his son: 7s - 2

In 3 years, the sum of their ages will be 52, In 3 years, the age of the sun will

increase 3, so s+3. Also, the age of his

father also increase 3, (7s-2)+3

(s+3)+ [(7s-2)+3] = 52 is the equation.

5 0

2 years ago

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