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1 year ago

8

Sam invested $7500, some at 7% interest and the rest at 10%. How much did he invest at each rate if he received $570.00 in inter

est in one year?

1 answer:

beks73 [17]1 year ago

8 0

Answer:

.

Step-by-step explanation:

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Does this Graph represent a function? Why or why not?

koban [17]

Answer: Yes

Step-by-step explanation:

To solve this, imagine you have a vertical line. You can use a ruler or something straight. Move the ruler across the screen. Does the ruler intersect 2 points at once? If so, then it is not a function. If it only intersects one point, then it is a function.

Always remember: a function cannot have two x's with different y-values.

Hope that helped!

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2 years ago

How do you calculate ratios

Marrrta [24]

Answer:

A link for a calculator that I use for ratios

Step-by-step explanation:

https://www.calculatorsoup.com/calculators/math/ratios.php

4 0

2 years ago

What is the value of x in the equation 4(2x 10) = 0? −5 −7 5 7

Alex73 [517]

4(2x+10)= 0
8x+40=0
subtract 40 from both sides
8x+40-40=0-40
8x=-40
x=-40/8
x=-5

4(2x-10)=0
8x-40=0
Add 40 from both sides
8x-40+40=0+40
8x=40
x=40/8
x=5

4 0

2 years ago

Read 2 more answers

The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and unive

gtnhenbr [62]

Answer: (0.8468, 0.8764)

Step-by-step explanation:

Formula to find the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = sample proportion.

z* = Critical value

n= Sample size.

Let p be the true proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.

Given : Sample size = 3597

Number of students believe that they will find a job immediately after graduation= 3099

Then, $\hat{p}=\dfrac{3099}{3597}\approx0.8616$

We know that , Critical value for 99% confidence interval = z*=2.576 (By z-table)

The 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation will be

$0.8616\pm(2.576)\sqrt{\dfrac{0.8616(1-0.8616)}{3597}}$

$0.8616\pm (2.576)\sqrt{0.0000331513594662}$

$\approx0.8616\pm0.0148\\\\=(0.8616-0.0148,\ 0.8616+0.0148)=(0.8468,\ 0.8764)$

Hence, the 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. = (0.8468, 0.8764)

4 0

2 years ago

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

2 years ago