2 numbers that mulitply to be negative 486 and add to be 9

Castel and Kali are selling wrapping paper for a school fundraiser. Customers can buy rolls of plain wrapping paper and rolls of

True [87]

Answer:

Plain is $7 and Holiday is $15

Step-by-step explanation:

Castel: 2p + 5h = 89

Kali: 9p + 10h = 213

Double Castel's: 4p + 10h = 178

Subtract fro Kali's 9p + 10h = 213

- 4p + 10h = 178

5p = 35

p = 7 Plain is $7

Substitute 7 into one of the equations and solve for h:

2(7) + 5h = 89

14 + 5h = 89

5h = 75

h = 15 Holiday is $15

6 0

2 years ago

At president Lincoln's inauguration in 1861, the newspaper headlines. stated there were about 30,000 people in attendance. if th

jek_recluse [69]

34,999 for the largest and 25,001 for the smallest, if you go to that 1 person extreme.

6 0

3 years ago

(CO 3) Sixty-five percent of products come off the line ready to ship to distributors. Your quality control department selects 1

gregori [183]

Answer:

a. 0.0056

b. 0.0846

c. 0.0255

d. 0.8487

Step-by-step explanation:

The missing question for the binomial distribution problem is to find the probability of:

a. Fewer than 4

b. Fewer than 6

c. Fewer than 5

d. Fewer than 10

The step by step solution is attached as an image.

5 0

3 years ago

The temperature in Anchorage, Alaska was -3°C in the afternoon. After the sun went down, the temperature dropped 6 degrees. Writ

Ira Lisetskai [31]

Answer:

-9

Step-by-step explanation:

-3 - 6

-3 + -6

-9

hope this helps!

7 0

3 years ago

A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has

ExtremeBDS [4]

Answer:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

$p_v =P(t_{(23)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=136$ represent the sample mean

$s=11$ represent the sample standard deviation

$n=24$ sample size

$\mu_o =140$ 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 140, the system of hypothesis would be:

Null hypothesis: $\mu \geq 140$

Alternative hypothesis: $\mu < 140$

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:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=24-1=23$

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

$p_v =P(t_{(23)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

4 0

3 years ago