What are the domain and range of the function?

For which function is 5 NOT an element of the range?

Darina [25.2K]

B. y = -2 + 4
y = 2 not 5

5 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ 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 check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago

What weight of dry substance is in 150g of a 3% substance solution? What weight of an 8% solution can we have with the same weig

Pavlova-9 [17]

The answers are: 4.5 g and 56.25 g respectively.

Since the first type of measurement in this question is weight or mass, I'll suppose that the percentage concentration is % mass/mass. For that type of concentration measurement, just multiply the percentage by the total mass to get the mass of the wanted material.
So 150 g * 3% = 150 g * 0.03 = 4.5g

For the 8% solution with the same amount of dry substance, use the ratio of percentages, multiplied by the mass of the first solution to get the wanted amount of new solution:
3/8 * 150 g = 56.35 g

5 0

3 years ago

In a housing estate the direct proportion of flats to houses is 2:5. If there are 80 flats, how many houses are there?

Kobotan [32]

$80 \div 2 = 40 \\ 40 \times 5 = 200$
there are 200 houses

4 0

3 years ago

Please help! I'm not the best at dividing with exponents!

Makovka662 [10]

MrBillDoesMath!

Answer: The first radio button choice 8/3 * m^8

Steps:

120 (m^ 16) / 45 (m^8) =

120/45 * (m^16)/(m^8) =

Simplify 120/45 (15 or 5 *3 divides each). Since base "m" is same in both terms simply subtract exponents.

8/3 ( m^(16 - 8)) =

8/3 * m^8

MrB

5 0

4 years ago