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Black_prince [1.1K]

3 years ago

15

A savings account earns 7% interest each year, compounded quarterly (4 times a year). If a person invests $300 and makes no furt

her deposits or withdrawals, what is the balance in the account after 5 years

1 answer:

cricket20 [7]3 years ago

4 0

Answer:

.

Step-by-step explanation:

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There are 18 girls and 12 boys in class. What is the probability that a new kid in senior class will be a boy?

valentinak56 [21]

There are 18 + 12 total students.

18 are girls and 12 are boys, since the total is 30 students, that's the amount of probable outcomes, and since there are 12 boys, that's the amount of favorable outcomes, thus

12/30 or 0.4, 0.4 * 100 is 40%.

5 0

3 years ago

Distribute 1/3(12y - 9)

allochka39001 [22]

3. 4y-3
because 1/3*12y - 1/3*9 = 4y - 3

7 0

3 years ago

Suppose point P (4,−9) is translated according to the rule (x,y) → ( x+3, y−2). What are the coordinates of P′? Explain.

larisa86 [58]

Answer:

P'(7 , -11)

Step-by-step explanation:

what this tells us

(x,y) → ( x+3, y−2)

Is that the value of x of the new point is going to take (x + 3) with respect to the value of x of the other point.

Is that the value of y of the new point is going to take (y - 2) with respect to the value of y of the other point.

P (4,−9)

x = 4

y = -9

P'( x+3, y−2)

x + 3 =

4 + 3 = 7

y - 2 =

-9 - 2 = -11

P'(7 , -11)

5 0

3 years ago

Find the range, median, and mode for each data set

ziro4ka [17]

50, 66, 73, 74, 75, 75, 76, 76, 81, 81

Remember, the range is when you subtract the lowest number from the highest number :)

Range : 81 - 50 = 31

Remember, the median is the middle number of an ordered data set from least to greatest.

Median : 75

Remember, the mode is the number that is on there the most.

Mode : 75, 76, 81 are all considered to be modes. However, if that is not one of the answer options, then your answer would be : There are not modes.

~Hope I helped!~

4 0

4 years ago

I need to know how to do the whole thing and understand it.

patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

$\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2$ $\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33$

Variance:

$s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667$ $s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244$

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

$H_0:\;\mu_A=\mu_B$

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

$H_a:\;\mu_A>\mu_B$

Test statistic (t):

The test statistic of a two-sample t-test is given by

$t=\frac{\bar{x}_A-\bar{x}_B}{s_p}$

Where sp is the pooled standard deviation given by

$\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}$ $t=\frac{2-3.33}{0.763}=-1.74$

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

6 0

1 year ago