If product is addition and factor is to multiply then what is division

If f(x)=x^2-7 and h(x)=7x-5, then f(h(2))=

zaharov [31]

We are given 3 equations. Using these equations we can find f(h(2)).

h(x)=7x-5
h(2)=7(2)-5
h(2)=14-5
h(2)=9

f(x)=x^2-7
f(h(2))=(7x-5)^2-7
we know the answer for h(2) by now, so we can substitute for that.
f(9)=9^2-7
f(9)=81-7
f(9)=74

so f(h(2))=74

5 0

3 years ago

5/8 of the staff are male. 5/12 of the staff works part time at the aquarium. What is the fraction of the staff being a female?

mestny [16]

Answer:3/8

Step-by-step explanation:

5/8 of the staff are male. The entire staff is represented by the fraction 8/8. Taking 5/8 from this, what we have left is

8/8-5/8 = 3/8 of the staff are female.

5 0

3 years ago

If the degree in the numerator is smaller than the degree in the denominator and there is no vertical

mote1985 [20]

The answer is false.

8 0

3 years ago

Read 2 more answers

A hospital uses cobalt-60 in its radiotherapy treatments for cancer patients. Cobalt-60 has a half-life of 7

NeTakaya

Answer:

We have 197 g of Co-60 after 18 months.

Step-by-step explanation:

We can use the decay equation.

$M_{f}=M_{i}e^{-\lambda t}$

Where:

M(f) and M(i) are the final and initial mass respectively
λ is the decay constant (ln(2)/t(1/2))
t(1/2) is the half-life of Co
t is the time at the final amount of m

$M_{f}=228e^{-\frac{ln(2)}{7} 1.5}$

$M_{f}=197\: g$

Therefore, we have 197 g of Co-60 after 18 months.

I hope it helps you!

4 0

3 years ago

A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o

Mandarinka [93]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 22 ounces

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

Let $\mu$ = average ounces of passion fruit juice used to fill each bottle.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 22 ounces {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight of the passion fruit juice = 21.54 ounces

$\sigma$ = population standard deviation = 1.38 ounce

n = sample of bottles = 65

So, test statistics = $\frac{21.54-22}{\frac{1.38}{\sqrt{65} } }$

= -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

6 0

3 years ago