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3 years ago

Which equation represents the line passing through the point (4, −5) that is parallel to the line x + 2y = 10?

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

4 0

Answer:

get photomath it will answer this

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9. The function h(t) = 150 - 4t represents the height of a person repelling down

Kisachek [45]

Answer:

30 feet

Step-by-step explanation:

Hello!

Half an hour is 30 minutes, so that means we are asked to find h(30).

Solve for h(30):

h(30) = 150 - 4(30) $\implies$ Substitute 30 for t
h(30) = 150 - 120 $\implies$ Simplify
h(30) = 30

The person would be 30 feet above the ground in half an hour.

8 0

2 years ago

Read 2 more answers

What's the distance between the two points , ( -3 , -4 ) and ( 4 , 3 ) ?

mixer [17]

Answer: the distance between the two points in Decimal Form is 9.899

3 0

3 years ago

Read 2 more answers

One triangle on a graph has a vertical side of 7 and a horizontal side of 12.

Alisiya [41]

If you are working out the diagonal line you need to use Pythagoras Theroum: c² = a²+b²

c² = 7²+12² = 193
√c² = c = √193 = 13.89

The length of the diagonal side is 13.89,
Hope this helps! :)

4 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

4 Stretch Your Thinking: Brian has some boxes of paper clips. Some boxes hold 10 clips and some boxes hold 100. He has some pape

rjkz [21]

Answer:02929

Step-by-step explanation:

Oofy

5 0

2 years ago