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Serggg [28]
3 years ago
14

10. Find the slope of a line perpendicular to the line containing (-8, 10) and (0,9).

Mathematics
1 answer:
Paladinen [302]3 years ago
3 0

the slope would be 8/1

your welcome

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If a plane may be named by three noncollinear points on that plane, plane abc may also be named ____.
Bond [772]
Plane acb
plane bac
plane bca
plane cab
plane cba
8 0
2 years ago
At a theme park, 35% of daily visitors ride the log ride. If on one particular day the park has 60,000 visitors, how many can be
Neko [114]
The answers is D 21,000because you have to do 60,000x.35=21,000
3 0
2 years ago
Read 2 more answers
Find the value of x.
Allisa [31]

A straight line is 180°, so you can do:

109 + ? = 180 (? is the top angle of the triangle)

? = 71°

Because the triangle is an isosceles triangle, the other angle is 71° too.

71 + 71 + ? = 180

? = 38

You subtract 38 with 90 to find the other angle.

90 - 38 = 52°

To find ∠2 you subtract 52 with 180.

180 - 52 = 128

∠2 = 128

128 = 11x + 18

110 = 11x

10 = x

5 0
2 years ago
12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that p = 0.12.

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so n = 18

This probability is:

P = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

There is a 34.60% probability that 0 or 1 of them is nearsighted.

5 0
3 years ago
A family pays for 50 MB of internet connectivity for downloads. When measuring their internet speed, a family member observed a
harkovskaia [24]

Answer:

its 70% of the speed they expected

4 0
2 years ago
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