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3 years ago

I need help with this?

Mathematics

1 answer:

Masja [62]3 years ago

4 0

since u know that there is 60 minutes in a hour do 60 times 3 and u will get 180.then do 180 divided by28 and u will get ur final answer which is 6 minutes and 12 seconds

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Estimating with percents: Isaac and his sister plan to surprise their parents by buying then each a pair of shoes. They agree th

morpeh [17]

for estimating round both the price of shoes and the percentage to round numbers

97.65 is close to 100, so use 100 for the price of shoes

68% is close to 70 so use 70%

then multiply 100 by 70 %,

100 *70% = 70 dollars

so Isaac would pay around 70 dollars

5 0

2 years ago

Read 2 more answers

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].

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Vlad [161]

(2x 6)(x-1)=0 happy to help :)

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Anon25 [30]

Answer:

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Step-by-step explanation:

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Neporo4naja [7]

Hey it Tya!

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