All categories

3 years ago

Camryn's investment is growing at a rate of

Mathematics

1 answer:

Marat540 [252]3 years ago

7 0

Answer:

So you take 6.4% and multiple it by $3,000 and add it each year :D, the final answer should be 58,000$

Step-by-step explanation:

I did the math you are welcome for the answer

You might be interested in

Jim had some dimes, quarters, and nickels. First, he counted only the dimes and quarters and found that he had 9 coins. Then, he

Olegator [25]

Answer:

$1.90

Step-by-step explanation:

Represent the numbers of nickels, dimes and quarters by n, d and q.

Then: d + q = 9, or q = 9 - d.

Also, n + d = 10

Lastly, n + d + q = 15.

Let's substitute 9 - d for q in the equation immediately above:

n + d + (9 - d) = 15, or n + 9 = 15. Then n must be 6.

In summary, Jim has 6 nickels, (10 - 6) dimes and (9 - 4) quarters, or:

6 nickels, 4 dimes and 5 quarters

Thus, he has 5($0.05) + 4($0.10) + 5($0.25), or

$0.25 + $0.40 + $1.25, or $1.90

7 0

3 years ago

What is the domain and range?

Viefleur [7K]

Answer:

Domain is your x max and Min, Range is your y max and Min

Step-by-step explanation:

Domain would be -5 to 8 or (-5,8)

range would be -3 to 2 or (-3,2)

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!! Number 2, and 4 please!! (YOU HAVE TO FIND THE SURFACE AREA OF THE PRISM)

Elenna [48]

The answer for 2 is 192. The answer for 4 is 672.

8 0

3 years ago

What is 28% of 14,350

xenn [34]

Percent means parts out of 100 so
28%=28/100 or 0.28
'of' can be translated as multiply
28/100 times 14,350=401800/100=4018 or
0.28 times 14,350=4018

5 0

3 years ago

Read 2 more answers

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago