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3 years ago

Help me !!!!!!!!!!!!!!!!

Mathematics

2 answers:

Ivan3 years ago

6 0

With what question(s) do you need Help With? All of them?

77julia77 [94]3 years ago

4 0

1. 2000 * 86/100 = 1720

2. $750 * 80/100 = $600

3. 120/160 * 100 = 75%

4. 66/12 * 100 = 550%

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Round 12.034 to the nearest tenth

vovangra [49]

The answer is 12. Thanks (;

5 0

3 years ago

Read 2 more answers

$265 worth of tickets were sold Adult tickets cost $6 Child tickets cost $2.50 • 57 tickets were sold. How many adult tickets we

BARSIC [14]

Answer:

35 adult tickets were sold

Step-by-step explanation:

If 35 adult tickets were sold, that would be $210

If 22 child tickets were sold, that would be $55

When you add them together, you get 57 tickets and $265.

(I'm sorry I don't know how to explain this but I hope it helped anyway)

5 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

On a particular production line,the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What

KengaRu [80]

Answer:

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

Step-by-step explanation:

Consider the provided information,

Let X is the number of defective bulbs.

Ten light bulbs are randomly selected.

The likelihood that a light bulb is defective is 5%.

Therefore sample size is = n = 10

Probability of a defective bulb = p = 0.05.

Therefore, q = 1 - p = 1 - 0.05 = 0.95

Mean of binomial random variable: $\mu=np$

Therefore, $\mu=10(0.05)=0.5$

Variance of binomial random variable: $\sigma^2=npq$

Therefore, $\sigma^2=10(0.05)(0.95)=0.475$

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

5 0

3 years ago

The sales tax in Kadisha's state is 5%. Kadisha says she computes a 15% tip by multiplying the tax shown on her bill by 3. For a

lapo4ka [179]

Explanation:

Kadisha's method works because 15% is 3 times 5%, and both amounts are computed on the amount of the bill before tax is added.

7 0

3 years ago