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2 years ago

What would my final balance be if I put $650 in the bank for 60 months with an interest rate of 6%?

1 answer:

5 0

It depends on the terms of the account.

If interest is compounded annually, 650*1.06^5 ≈ 869.85 . . . . dollars.
If interest is compounded quarterly, 650*1.015^20 ≈ 875.46 . . dollars.
If interest is compounded monthly, 650*1.005^60 ≈ 876.75 . . .dollars.

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BRIANILEST!!!!Which lines are parallel? Check all that apply.

spin [16.1K]

Answer:

b║f

c║e

Step-by-step explanation:

Given a diagram,

Lines b, c, d, e, f and a transversal line 'a'.

Here line 'b' makes an angle with line 'a' is 89.5° and line 'f' makes an angle with line 'a' is also 89.5° (180°-90.5° = 89.5°)

Since corresponding angles are same.

So, line 'b' is parallel to line 'f'.

And line 'c' makes an angle with line 'a' is 91.5° and line 'e' makes an angle with line 'a' is also 91.5° (180°-88.5° = 91.5°)

Since corresponding angles are same.

So, line 'c' is parallel to line 'e'.

Therefore,

b║f

c║e

That's the final answer.

4 0

3 years ago

Read 2 more answers

Harish

Artist 52 [7]

Answer:

Step-by-step explanation:

Distance traveled in $1\frac{4}{5}$ hours = $\frac{3}{10}$

Distance traveled in 1 hour = $\frac{3}{10}$ ÷ $1\frac{4}{5}$

$=\frac{3}{10}$ ÷ $\frac{9}{5}$

$= \frac{3}{10}*\frac{5}{9}\\\\=\frac{1}{2}*\frac{1}{3}\\\\=\frac{1}{6}$

distance traveled in 3 1/5 hours = $\frac{1}{6}*3\frac{1}{5}$

= $=\frac{1}{6}*\frac{8}{5}\\\\=\frac{1}{3}*\frac{4}{5}\\\\=\frac{4}{15}$

5 0

2 years ago

Solve for x If you don’t know how to do it don’t answer

goblinko [34]

Your giving 5 points for this shi

3 0

2 years ago

Can you plzzzzzzzz help meeeee it is urgent for the 1st and second question

kherson [118]

Answer:

n<-9

p (put the symbol already there here) -7

Step-by-step explanation:

Hope this helps!!

7 0

2 years ago

If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e

Pepsi [2]

The outlier of a dataset is a data element that is relatively far from the remaining data elements

99 is an outlier of pet group
See attachment for the parallel box plots

(a) Prove that 99 is an outlier for Pet

We have:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

$n = 15$

The quartiles positions are:

$Q_1 = \frac{n + 1}{4}$

$Q_1 = \frac{15 + 1}{4}$

$Q_1 = \frac{16}{4}$

$Q_1 = 4th$

$Q_3 = Q_1 \times 3$

$Q_3 = 4th \times 3$

$Q_3 = 12th$

So, we have:

$Q_1 = 4th$

$Q_3 = 12th$

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

$Q_1 = 68$

$Q_3= 79$

The lower and upper limits of the outlier are:

$L = Q_1 - 1.5 \times (Q_3 - Q_1)$

$U = Q_3 + 1.5 \times (Q_3 - Q_1)$

So, we have:

$L = 68 - 1.5 \times (79 - 68)$

$L = 51.5$

$U = 79+ 1.5 \times (79 - 68)$

$U = 95.5$

This means that data below 51.5 or above 95.5 are outliers.

Hence, 99 is an outlier because 99 is greater than 95.5

(b) The parallel box plot

The three groups are:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82

Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0

2 years ago