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Bumek [7]
3 years ago
6

300 is what percent of 150?.

Mathematics
2 answers:
stepan [7]3 years ago
6 0
300 : x = 150 : 100
x = 300 * 100 : 150
x = 200
------------
200%
nadya68 [22]3 years ago
4 0
We can use algebra to solve this.
Let the percentage be P%.
From the given information,
150 x P% = 300
Flip x150 to the other side and turn it into ÷150
P% = 300 / 150
P% = 2
We can turn a whole number into decimal form by moving the decimal place of it 2 places to the right.
In 2, the decimal is behind 2, which makes it 2.0
So when we move the decimal place to 2 places to the right,
The answer would be 200%
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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
3 years ago
What are the factors of 45?
podryga [215]
Factors: Numbers that can multiply into a certain number.

The factors of 45 are:
1 (x 45)
3 (x 15)
5 (x 9)
9 (x 5)
15 (x 3)
1 (x 45)


4 0
3 years ago
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3 years ago
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8 0
3 years ago
You were given the polynomial f(x)=3x^3+13x^2+19x+k and you know that f(-2)=0find k
eduard

Answer:

k = 10

Explanation:

The initial polynomial is:

f(x)=3x^3+13x^2+19x+k

If f(-2) = 0, then when we replace x by -2, the result will be 0. It means that we can write the following equation:

\begin{gathered} f(-2)=3(-2)^3+13(-2)^2+19(-2)+k=0 \\ 3(-2)^3+13(-2)^2+19(-2)+k=0 \end{gathered}

Therefore, we can solve for k as follows:

\begin{gathered} 3(-8)+13(4)+19(-2)+k=0 \\ -24+52-38+k=0 \\ -10+k=0 \\ -10+k+10=0+10 \\ k=10 \end{gathered}

So, the value of k is 10

6 0
1 year ago
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