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ira [324]
3 years ago
13

A lab is trying to determine if a new medication is effective at reducing acne breakouts. The results are displayed in the Venn

Diagram below:What is the probability that the person's skin cleared up given that they used the medication?
A)10%
B)30%
C)20%
D)60%

Mathematics
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

its %60 i just took the test and got it right

Step-by-step explanation:

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HELP PLEASE WILL GIVE BRAINLIEST BE QUICK
slamgirl [31]

Answer:

(2,0) and (4,0)

Step-by-step explanation:

The roots of the parabola are the points where the value of the function is zero, i.e., where the graph crosses the x-axis.

(2,0) and (4,0)

7 0
2 years ago
Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th
bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0
3 years ago
What to numbers multiply to -38 and add to 17
Ymorist [56]

Answer:

-2 and 19

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The ten students in a club are lined up in a row for a group photograph. How many different arrangements are possible if the clu
kondor19780726 [428]

We know that

if the club includes one set of identical triplets wearing matching clothes

then

the number of different arrangements that are possible is

10! / 3! = (10*9*8*7*6*5*4*3!)/3!

=604,800

The answer is

<span>604,800</span>

8 0
2 years ago
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Question 1 of 10
sergiy2304 [10]

Answer: C. 2 and 3 only

Step-by-step explanation:  i did the same quiz

6 0
2 years ago
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