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2 years ago

Which function has the graph shown below? *

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

7 0

Answer:

Step-by-step explanation:

yeah-ya......... right?

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Choose the graph below that represents the following system of inequalities:

castortr0y [4]

Yshejrhhe is they easiest of them all and as you can see I just can’t fail it bro ik too smart yk it’s y>-3x +1-5

7 0

1 year ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

8 0

3 years ago

How many ways are there to select a 5-card hand from a regular deck such that the hand contains at least one card from each suit

Snezhnost [94]

Answer:

There are 685464 ways of selecting the 5-card hand

Step-by-step explanation:

Since the hand has 5 cards and there should be at least 1 card for each suit, then there should be 3 suits that appear once in the hand, and one suit that apperas twice.

In order to create a possible hand, first we select the suit that will appear twice. There are 4 possibilities for this. For that suit, we select the 2 cards that appear with the respective suit. Since there are 13 cards for each suit, then we have ${13 \choose 2} = 78$ possibilities. Then we pick one card of all remaining 3 suits. We have 13 ways to pick a card in each case.