△ABC is reflected over the x−axis and then reflected over the y−axis to form ΔA''B''C'' as shown below.

Kelvin and Lewie each design surveys in order to determine the average number of people who buy food at the mall. Kelvin surveys

Lisa [10]

Answer:

Lewie Survey produces less valid result.

Step-by-step explanation:

Given:

1. Kelvin surveys every other person leaving the food area.

2. Lewie surveys every fifth person leaving the mall's main entrance.

Therefore,

Kelvin will able to produce valid result as he is surveying every person who is leaving food court.

Means,

There are high chances that would have bought food from the court.

Whereas,

Lewie is surveying every fifth person leaving the the mall's entrance.

Therefore,

Here probability to produce valid result is less as they may have not buy food or they may visited other shops.

So, Lewie survey will produce less valid result.

8 0

2 years ago

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From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that ag

Jlenok [28]

Answer:

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

Step-by-step explanation:

1) Data given and notation

$\bar X=18.1$ represent the mean age when smokers first start to smoke varies

$s=1.3$ represent the standard deviation for the sample

$\sigma=2.1$ represent the population standard deviation

$n=37$ sample size

$\mu_o =5.7$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean starting age is at least 19, the system of hypothesis would be:

Null hypothesis: $\mu\geq 19$

Alternative hypothesis: $\mu < 19$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{18.1-19}{\frac{2.1}{\sqrt{37}}}=-2.607$

Calculate the P-value

Since is a one-side lower test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

4 0

2 years ago

Which rule should be applied to reflect f(x)=x^3 over the y-axis?

Anna11 [10]

To make a reflection over the y axis, make the whole equation negative.
The new equation should look like this:

f(x) = -x^3

7 0

2 years ago

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A recent survey shows that the probability of a college student drinking alcohol is 0.6. further, given that the student is over

Murrr4er [49]

Answer:

0.24

Step-by-step explanation:

These events are not mutually exclusive; this means they can happen at the same time.

For two events A and B that are not mutually exclusive,

P(A and B) = P(A) * P(B|A)

Let A be the event "over 21 years old" and B be the event "drinks alcohol".

The probability that a student is over 21 years old is 0.3; this is because 30% of the students are over 21 years old.

The probability that a student drinks alcohol given they are over 21 is 0.8.

This gives us

P(A and B) = 0.3(0.8) = 0.24

8 0

2 years ago

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Nathan had an infection, and his doctor wanted him to take penicillin. Because Nathan’s father and paternal grandfather were all

Sergio039 [100]

Let events
A=Nathan has allergy
~A=Nathan does not have allergy
T=Nathan tests positive
~T=Nathan does not test positive

We are given
P(A)=0.75 [ probability that Nathan is allergic ]
P(T|A)=0.98 [probability of testing positive given Nathan is allergic to Penicillin]

We want to calculate probability that Nathan is allergic AND tests positive
P(T n A)

From definition of conditional probability,
P(T|A)=P(T n A)/P(A)
substitute known values,
0.98 = P(T n A) / 0.75
solving for P(T n A)
P(T n A) = 0.75*0.98 = 0.735

Hope this helps!!

3 0

2 years ago