If you know that a relationship is proportional and are given one ordered pair other than (0,0), how can you find another pair?

Which of the following can not be used to calculate 32% of 50

stepladder [879]

Answer:

the answer is 16

Step-by-step explanation:

6 0

3 years ago

Whats the answer and how

Jet001 [13]

Answer: its c or 35

Step-by-step explanation: because at 10 salespeople its 40, so 9 salespeople is 35

8 0

3 years ago

How many different sums can be made with a penny nickel dime quarter?

weqwewe [10]

1 coin is 4 ways
2 coins is 4*3 ways
3 coins is 4*3*2 ways
4 coins is 4*3*2*1 ways

so total is 4+12+24+24=64 sums

8 0

3 years ago

A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses p

gizmo_the_mogwai [7]

Answer:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

$\bar X_{1}=9.1$ represent the mean for the sample 1

$\bar X_{2}=8$ represent the mean for the sample 2

$s_{1}=1.9$ represent the sample standard deviation for the sample 1

$s_{2}=2.1$ represent the sample standard deviation for the sample 2

$n_{1}=40$ sample size for the group 1

$n_{2}=50$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

Statistical decision

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

$df=n_{1}+n_{2}-2=40+50-2=88$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

3 0

3 years ago

Polygon PQRS, shown in the figure, is dilated by a scale factor of 0.5 with the origin as the center of dilation, resulting in p

nata0808 [166]

Based on the above:

The slope of P'Q' is = -3/2

The length of P'Q' is approximately = $\sqrt[3]{13}$