Answer pls? Bias matching??

HELP PLEASE FOR BRAINLEST

Semenov [28]

Answer:

3/13

Step-by-step explanation:

Put the total number that tapped the ground AND 5/10 worms (which is 3) over the total number which is 13

so 3/13 which doesn't simplify

hope this helped :)

5 0

2 years ago

I NEED HELP WITH THIS ASAP I CAN NOT FIND ANY COMMON GROUND FOR THESE NUMBERS ALL HELP IS APPRECIATED THANK YOU

poizon [28]

Answer:

45 is um 9x+xy-4y

Step-by-step explanation:

I did it it was a simplify type thing

3 0

2 years ago

Write 343 with an exponent by using 7 as the base

pochemuha

7^3 = 343

7×7=48
7×7×7=343

8 0

3 years ago

Read 2 more answers

If 20% of the people in a community use the emergency room at a hospital in one year, find

Pie

Answer:

a) 87.91% probability that at most three used the emergency room

b) 20.13% probability that exactly three used the emergency room.

c) 3.28% probability that at least five used the emergency room

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use the emergency room, or they do not. The probability of a person using the emergency room is independent of any other person. So we use the binomial probability distribution to solve this question.

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 combinations 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.

Sample of 10 people:

This means that $n = 10$

20% of the people in a community use the emergency room at a hospital in one year

This means that $p = 0.2$

a) At most three used the emergency room

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$

87.91% probability that at most three used the emergency room

b) Exactly three used the emergency room

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

20.13% probability that exactly three used the emergency room.

c) At least five used the emergency room

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

From 0 to 3, we already have in a).

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1074 + 0.2684 + 0.3020 + 0.2013 + 0.0881 = 0.9672$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9672 = 0.0328$

3.28% probability that at least five used the emergency room

4 0

3 years ago

Suppose you are testing the null hypothesis that a population mean is less than or equal to 46, against the alternative hypothes

PolarNik [594]

Answer:

Option A) 2.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 46

Sample mean, $\bar{x}$ = 50

Sample size, n = 25

Alpha, α = 0.05

Population standard deviation, σ = 8

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 46\\H_A: \mu > 46$

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{50 - 46}{\frac{8}{\sqrt{25}} } =2.5$

Option A) 2.5

6 0

3 years ago