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IRISSAK [1]
3 years ago
6

Write 9/11 as an expression

Mathematics
2 answers:
Anastasy [175]3 years ago
7 0
9 divided by 11 the / is basically a division sign.
podryga [215]3 years ago
4 0
It is 9 divided by 11 wich is 0.8181818182
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NEED HELP ASAP PLEASE
Ilya [14]

Answer:

see explanation

Step-by-step explanation:

Given

A = \frac{1}{2} bh

a. substitute b = 6 and h = 5 into the formula

A = \frac{1}{2} × 6 × 5 = \frac{1}{2} × 30 = 15 units²

b.

Multiply both sides of the formula by 2

2A = bh, that is

bh = 2A ← divide both sides by b

h = \frac{2A}{b} = \frac{2(16)}{4} = \frac{32}{4} = 8

c

Using

bh = 2A ← divide both sides by h

b = \frac{2A}{h} = \frac{2(12)}{4} = \frac{24}{4} = 6

3 0
3 years ago
(-2,17)(-1,13)(1,5)(2,1)(3,-3)<br> The rate of change is ____.
7nadin3 [17]
Answer:

The rate of change in -4

Step-by-step explanation:

The rate of change is also known as the slope

To do this, take two points {(-2, 17) , (-1, 13)} and find the slope

m = slope

m = change in y/change in x

m = 13 - 17/-1 - (-2)

m = -4/1

m = -4

The rate of change in -4
4 0
2 years ago
How much would you need to deposit in an account now in order to have $6000 in the account in 5 years
777dan777 [17]

Answer:

5 years=$10

Step-by-step explanation:

count by 5's

5 0
3 years ago
A student found a strong correlation between the age of people who run marathons and their marathon time. Can the student say th
DENIUS [597]
Elite marathon race times improved from 5 to ~20 years, remained linear between ~20 and ~35 years, and started to increase at the age of ~35 years in a curvilinear manner with increasing age in both women and men. The sex difference in elite marathon race time increased non-linearly and was lowest at the age of ~49 years.
7 0
3 years ago
Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25
Murrr4er [49]

Answer:

The margin of error  for a 90% confidence interval is 16.4

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64

Margin of error =

z_{critical}\times \dfrac{\sigma}{\sqrt{n}}

Putting the values, we get,

1.64\times \dfrac{50}{\sqrt{25}} = 16.4

Thus, the margin of error  for a 90% confidence interval is 16.4

8 0
3 years ago
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