liz earns $400 for 40 hours of work.use a ratio table to determine how much she earns for 6 hours of work

zloy xaker [14]

Answer:

A

Step-by-step explanation:

This is the answer because the others don't make sense to me at all...

PLEASE DO MARK ME AS BRAINLIEST UWU 

3 0

3 years ago

Martina chose a random sample of 10 bags of candy from two different Brands, A and B. All bags in the sample are the same size.

Alborosie

I found the dot plots that accompanies this problem.

Based on the plots, the statement that gives is a valid comparison of the number of candies in the bags of the two Brands is:

B. The number of candies in the bags from Brand B is greater and less consistent than the number of candies in the bags from Brand A.

Dots in Brand B are scattered and whereas dots in Brand A are not and they are more concentrated between 52 to 55 range.

7 0

3 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

Which quantity is larger?

Kisachek [45]

Remember
$(x^m)^n=x^{mn}$
convert to base 10
$100^{500}=(10^2)^{500}=10^{1000}$
$1000^{300}=(10^3)^{300}=10^{900}$

$10^{1000}\ \textgreater \ 10^{900}$

A is larger

8 0

3 years ago

Read 2 more answers

What is the distance to the Earth's Horizon from point P? Enter enter the decimal in the Box round only your final answer to the

Tasya [4]

Answer:

<h2>The distance to the Eath's Horizon from point P is 352.8 mi, approximately.</h2>

Step-by-step explanation:

You observe the problem from a graphical perspective with the image attached.

Notice that side $x$ is tangent to the circle, which means is perpendicular to the radius which is equal to 3,959 mi.

We have a right triangle, that means we need to use the Pythagorean's Theorem, to find the distance to the Earth's Horizon from point P.

The hypothenuse is 3959 + 15.6 = 3974.6 mi.

$(3974.6)^{2}=x^{2} +(3959)^{2} \\x^{2} =15,797,445.16 - 15,673,681\\x=\sqrt{123,764.16} \approx 351.8$

Therefore, the distance to the Eath's Horizon from point P is 352.8 mi, approximately.

4 0

4 years ago