What is the difference of 21 1/4

Helpppppppppp plzzzzzz

stiks02 [169]

Answer:

its D

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Solve In (2x+3)=7 round to the nearest thousandth

e-lub [12.9K]

To solve this problem you must apply the proccedure shown below:

1. You have:

In(2x+3)=7

2. Then, you must apply log(e), as below:

In(2x+3)=ln(e^7)

3. Now, you obtain:

2x+3=e^7

4. Youy must clear the variable "x", as below:

2x=e^7-3
 x=(e^7-3)/2

5. Therefore, the value of "x" is:

x=546.817

The answer is: x=546.817

8 0

2 years ago

150% of what number is 40.

Crazy boy [7]

Solution for what is 150% of 40 40/x=100/150 (40/x)*x=(100/150)*x - we multiply both sides of the equation by x 40=0.666666666667*x - we divide both sides of the equation by (0.666666666667) to get x 40/0.666666666667=x 60=x x=60 now we have: 150% of 40=60

8 0

2 years ago

Kay's mother taught her how to make handmade ornaments to sell at a craft fair. Kay rented a table at the fair for $30 and set u

leva [86]

Answer:

profit= 6. 00x - (30+2. 50x)

5 0

2 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

2 years ago

What is the difference of 21 1/4 - 18 2/4​

What is the difference of 21 1/4 - 18 2/4