Your used car payment will be $250 per month. What financial challenge do you now face?

A right rectangle prism has a length of 5 cm, a width of 4 cm, and a height of of 3 cm. The dimensions of the prism are doubled.

MaRussiya [10]

As the volume of cuboid is given by,
l×b×h
as the dimensions are doubled,
l2×b2×h2
5×2×4×2×3×2
= 480 cm^3

6 0

3 years ago

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What is 40% of $910?

borishaifa [10]

910*0.4=364

So the anwser is $ 364

8 0

3 years ago

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<img src="https://tex.z-dn.net/?f=3%20%2B%20%284%20%2B%205%29%20%3D%20%283%20%2B%204%29%20%2B%20m" id="TexFormula1" title="3 + (

max2010maxim [7]

Answer:

The value of m in the expression is 5

Step-by-step explanation:

Given expression as :

3 + ( 4 + 5 ) = ( 3 + 4 ) + m

Or, 3 + ( 9 ) = ( 7 ) + m

Or, 12 = 7 + m

∴ m = 12 - 7 = 5

Hence The value of m in the expression is 5 Answer

8 0

3 years ago

Jackson bought 3 pounds of candies for $9.75 What was the price of these candies in dollars per pound?

Stels [109]

9.75 dollars divided by 3 pounds equals 3.25 per pound

8 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

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

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago