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3 years ago

I need help! Please!

Mathematics

1 answer:

lakkis [162]3 years ago

5 0

134,217,728/16,777,216

Even more simplified dividing them would equal 8

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Which ones do I choose?

Snezhnost [94]

Answer:

-6w-22

Step-by-step explanation:

-3(2w+6)-4=

= -6w-18-4=

= -6w-22

So,choose C (none of the above) :)

7 0

2 years ago

HELP NEEDED!!!

cluponka [151]

Answer:

Step-by-step explanation:

4 0

3 years ago

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The sum of the lengths of two opposite sides of the circumscribed quadrilateral is 12 cm, the length of a radius of the circle i

nasty-shy [4]

Answer:

60cm^2

Step-by-step explanation:

We assume that is a circumscribing quadrilateral, rather than one that is circumscribed. It is also called a "tangential quadrilateral" and its area is ...

K = sr

where s is the semi-perimeter, the sum of opposite sides, and r is the radius of the incircle.

K = (12 cm) (5cm) = 60 cm²

_____

A quadrilateral can only be tangential if pairs of opposite sides add to the same length. Hence the given sum is the semiperimeter.

8 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

2 years ago

A2 + 7a + 10 = (a + 5)(a + ? )

Brilliant_brown [7]

$a^{2}$ +7a+10 = (a+5)(a+2)

3 0

3 years ago

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