1. Round to the nearest hundredth.

8. Bella has 6.3 kilograms of berries. She packs 0.35 kilogram of berries into each

vovangra [49]

Answer:

If my math is correct it should be $53.82

Step-by-step explanation:

You Divide 6.3kg by 0.35kg that would give you 18. Then you multiply $2.99 by 18 and that should get you $53.82

7 0

3 years ago

FIND THE SLOPE OF THE LINE... A) 1/2 B) 2 C) - 1/2 D) -2

Ainat [17]

Answer:

The slope would be C) -1/2

Step-by-step explanation:

To find it, we need to use the slope formula and two points on the line. The points we'll use are (0, -3) and (2, -4)

m(slope) = (y2 - y1)/(x2 - x1)

m = (-3 - -4)/(0 - 2)

m = 1/-2

m = -1/2

8 0

3 years ago

Standard deviation of 19,29,21,22,23

MaRussiya [10]

Answer:

3.370459909

Step-by-step explanation:

Figure out the arithmetic mean of the numbers (22.8)

Subtract the mean from each number and square the difference

Add up the squared differences and divide the number by 5

Square root answer

3 0

3 years ago

Which polynomial represents the area of the rectangle? 3x-5= x+1

maxonik [38]

Answer: 3x² - 2x - 5

Step-by-step explanation:

The area of a rectangle is simply length * width. Thus, simply multiply them, and FOIL.

(3x - 5) (x + 1)

3x ^2 + 3x - 5x - 5

Combine like terms

3x² - 2x - 5

Hope it helps :) and let me know if you want me to elaborate.

4 0

3 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

3 years ago