What is the domain and range of the function?

Travis and Paula went to lunch. Travis ordered a sandwich for $7.50, and Paula ordered a burger for $5.25. After lunch, they lef

Katen [24]

TRAVIS AND PAULA SPEND FOR LUNCH 7.50+5.25=12.75 DOLLARS
WE NEED TO FIND OUT HOW MUCH IS 1% OUT OF 12.75 WHICH IS
12.75/100=.1275
.1275*15=1.91 TIP LEFT
12.75+1.91=14.66

3 0

3 years ago

Read 2 more answers

bill and sam have a coupon for a pizza with 1 topping. The choices of toppings are: pepperoni, sausage, mushrooms, green peppers

cricket20 [7]

Well there are 6 toppings. For one person to select sausage, it is $\frac{1}{6}$ . For two people, multiply them together and the probability is $\frac{1}{36}$

5 0

3 years ago

Plz help me!!!!

sveta [45]

A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.

b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed.

4 0

3 years ago

15•(3 + b) = 45 + b

Alekssandra [29.7K]

Subtract 15 from both sides that gets rid of 15 and your left with 3b =30b divide 3 on both sides and the answer is 10

8 0

3 years ago

Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving

Morgarella [4.7K]

Answer:

a)

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3

b) 34%

c) practically 0

d) Reject the null hypothesis.

Step-by-step explanation:

a)

Since an individual aged 18 to 32 either continues to be dependent on their parents or not, this situation follows a Binomial Distribution and, according to the previous research, the probability p of “success” (depend on their parents) is 0.3 (30%) and the probability of failure q = 0.7

According to the sample, p seems to be 0.34 and q=0.66

To see if we can approximate this distribution with a Normal one, we must check that is not too skewed; this can be done by checking that np ≥ 5 and nq ≥ 5, where n is the sample size (400), which is evident.

We can then, approximate our Binomial with a Normal with mean

$\bf np = 400*0.34 = 136$

and standard deviation

$\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742$

Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3

So, this is a right-tailed hypothesis testing.

b)

According to the sample the proportion of "millennials" that are continuing to be dependent on their parents is 0.34 or 34%

c)

Our level of significance is 0.05, so we are looking for a value $\bf Z^*$ such that the area under the Normal curve to the right of $\bf Z^*$ is ≤ 0.05

This value can be found by using a table or the computer and is $\bf Z^*$ = 1.645

Applying the continuity correction factor (this should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal)), we simply add 0.5 to this value and

$\bf Z^*$ corrected is 2.145

Now we compute the z-score corresponding to the sample

$\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}$