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3 years ago

How do i find the decumal equivalent for 2 over 3

1 answer:

3 0

To find the decimal equivalent for 2/3, you simply divide 2 by 3.
$2 \div 3$
After dividing, the decimal equivalent will be about 0.6666666667

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At a summer camp, there are 15 counselors, 5 cooks, and 120 campers. 5 points

lesya [120]

Answer:

eto 1:12 pero di ko sure,antayin mo nlng yung sagot ng iba

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3 years ago

Amelia made 3 1/3 pounds of trail mix. If she puts 2/3 of a pound into each bag, how many bags can Amelia fill? Write your answe

Neporo4naja [7]

Answer:

Step-by-step explanation:

Because ⅔ goes into 3 ⅓ 6 times.

7 0

3 years ago

Workout the area of the quarter of a circle with a radius of 8

QveST [7]

Answer:

$16\pi$

Step-by-step explanation:

We need to find 1/4 of the area of a circle with a radius of 8

Let r be radius

Area of a circle = $r^{2} \pi$

Substitute r with 8

Area of a circle = $8^2\pi =64\pi$

Now, to find 1/4 of the area, we can create an equation $\frac{64\pi }{4}={16\pi$

The answer is $16\pi$

Hope this helps :)

Have a spectacular day!

3 0

3 years ago

Read 2 more answers

There are 125 members in the school

Igoryamba

Answer:

125x + 5,000 > 25,000

5 0

3 years ago

Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a

krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. <u>Finding z-score for Reagan;</u>

Let X = distribution of SAT scores

SO, X ~ Normal( $\mu=1000, \sigma^{2}=100^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1000

$\sigma$ = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = $\frac{1140-1000}{100}$ = 1.4

2. <u>Finding z-score for Jessie;</u>

Let X = distribution of ACT scores

SO, X ~ Normal( $\mu=17, \sigma^{2}=5^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 17

$\sigma$ = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = $\frac{30-17}{5}$ = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

6 0

3 years ago