All categories

2 years ago

If anyone understands can you please help me✨‍♀️

Mathematics

1 answer:

sergiy2304 [10]2 years ago

4 0

180 mins to walk 12 miles :)

You might be interested in

Find the area of each segment, round to the nearest tenth

FrozenT [24]

Answer:

Idk

Step-by-step explanation:

8 0

3 years ago

What is the formula for calculating a residual value

Ksivusya [100]

The formula to figure residual value follows: Residual Value = The percent of the cost you are able to recover from the sale of an item x The original cost of the item. For example, if you purchased a $1,000 item and you were able to recover 10% of its cost when you sold it the residual value is $100!!

7 0

2 years ago

Read 2 more answers

What is equivalent to (3 - 2x) (4x - 7)

bonufazy [111]

Answer:

-8x^2+26x-21

Step-by-step explanation:

(3-2x)(4x-7)

12x-8x^2-21+14x

-8x^2+12x+14x-21

-8x^2+26x-21

7 0

2 years ago

Read 2 more answers

A van can travel 18 miles on each gallon of gasoline. at that rate how many miles can the van travel on 15 gallons of gasoline

tensa zangetsu [6.8K]

18 / 1 = x / 15......18 miles to 1 gallon = x miles to 15 gallon
cross multiply
(1)(x) = (18)(15)
x = 270 miles

7 0

2 years ago

Read 2 more answers

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