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2 years ago

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Average speed of Car 1 = 45 mph. Average speed of Car 2 = 65 mph. Time elapsed between start of Car 1 and start of Car 2 = 18 mi

nutes. How long before Car 2 overtakes Car 1? hour.

1 answer:

Art [367]2 years ago

3 0

Is this math? if so let me know maybe I could try to help

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your friend deposits $9500 in an investment account that earns 2.1% annual interest. what is the balance after 11 years when the

solmaris [256]

Answer:

The answer is "$ 11,961.43"

Step-by-step explanation:

Given values:

P = $ 9500

r= 2.1 %

time (t)= 11 year

total quarterly year =4

Formula:

$A= P (1+\frac{r}{n})^{nt}\\\\$

quarterly time = $4 \times 11$

$= 44$

$A= P (1+\frac{r}{n})^{nt}\\\\$

$A = 9500 ( 1+ \frac{0.021}{4})^{44}\\\\A = 9500(\frac{4+ 0.021}{4} )^{44} \\\\A = 9500(\frac{4.021}{4} )^{44} \\\\A= 9500 (1.00525)^{44}\\\\A= 9500(1.25909819)\\\\A= 11,961. 43\\$

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2 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

2 years ago

Two bakeries sell muffins that can be customized with different kinds of berries. Berry Bakery sells muffins for $14.50 a dozen,

vfiekz [6]

Yes. The original price of the dozen added to the price of berries (.5) multiplied by the amount(b) is the same equational representation to the right of the equal sign.

5 0

2 years ago

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A manufacturer produces two models of toy airplanes. It takes the manufacturer 20 minutes to assemble model A and 10 minutes to

Georgia [21]

Answer:

A - 90 units

B = 0 units

Step-by-step explanation:

Here we have two models A and B with the following particulars

Model A B (in minutes)

Assembly 20 15

Packing 10 12

Objective function to maxmize is the total profit

$z=12A+8B$ where A and B denote the number of units produced by corresponding models.

Constraints are

$20A+15B\leq 1800\\\\10A+12B\leq 1080$

These equations would have solutions as positive only

Intersection of these would be at the point

i) (A,B) = (60,40)

Or if one model is made 0 then the points would be

ii) (A,B) = (90,0) oriii) (0, 90)

Let us calculate Z for these three points

A B Profit

60 40 1040

90 0 1080

0 90 720

So we find that optimum solution is

A -90 units and B = 0 units.

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2 years ago

Noah finds an expression for V(x) that gives the volume of an open-top box in cubic inches in terms of the length x in inches of

Fudgin [204]

Answer:

its -64937.4.5.53.94x

Step-by-step explanation:

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2 years ago

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