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3 years ago

Plss Help!!!.

Mathematics

1 answer:

RideAnS [48]3 years ago

7 0

Answer:

$ 300.

Step-by-step explanation:

60% of 500 = $\frac{60}{100}*500$

= 60 * 5

= $ 300

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The prices of karaoke at 6 different stores are $77, $85, $78, $72, $80 and $118. What is the outlier in tbe data set?

julsineya [31]

$118 is the outliner because it is the furthest away in price compared to the rest.

8 0

3 years ago

What’s the number between -6 and -7 on a number line

CaHeK987 [17]

Answer: -6.5

Step-by-step explanation:

To find the number between them,find their average or mean .

-6 + -7 = -13

-13/2 = -6.5

The number between them is -6.5

7 0

3 years ago

How does the suffix help you understand the meaning of announcement

sergeinik [125]

The suffix 'ment' belongs to nouns that denote an action. the word 'announce' means to declare or make something known. So the word 'announcement' is the action of declaring.

6 0

3 years ago

PLEASE HELP ASAP!!!! WORTH 15 POINTS!

Finger [1]

Answer:

Its C I know but I have my friend that knows this by heart

3 0

3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago