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3 years ago

14

HELP ASAP!

2 answers:

OleMash [197]3 years ago

8 0

Answer:

Both Kurt and Maria are right, because 3,585 should be rounded up to 3,600, and 21% of this amount can be approximated by either finding 10% of 3,600 and multiplying the result by 2 or by finding 1/5 of 3,600.

Step-by-step explanation:

Umnica [9.8K]3 years ago

3 0

D. Both Kurt and Maria are right, because 3,585 should be rounded up to 3,600, and 21% of this amount can be approximated by either finding 10% of 3,600 and multiplying the result by 2 or by finding 1/5 of 3,600.

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Evaluate the expression<br> I 13 I + I -2 I

sertanlavr [38]

Answer:

15

Step-by-step explanation:

Apply absolute rule: | a | = a, a ≥ 0

| 13 | = 13

= 13 + | -2 |

| -2 | = 2

= 13 + 2

= 15

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3 years ago

Where can we put parentheses in 2 x 3+1+2 to make it equivalent to 12?

zhannawk [14.2K]

2(3+1+2) is equivalent to 12

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4 years ago

What is 0.000000685 written in scientific notation?

maria [59]

we have

$0.000000685$

we know that

$0.000000685=\frac{685}{1,000,000,000} =\frac{685}{10^{9}}=685*10^{-9}$

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therefore

<u>the answer is</u>

$6.85*10^{-7}$

3 0

3 years ago

Read 2 more answers

Molecules of a toxic chemical eventually decompose into inert substances. Suppose the decomposition time is exponentially distri

koban [17]

Answer:

a) 4.16 years

b) 27.73 years

c) 238.44 years

d) 3,870.53 years

Step-by-step explanation:

Let X be the random variable that measures the decomposition time.

a)

$\bf \lambda =6$

In this case, since the decomposition time is exponentially distributed with a mean of 1/6, we have

$\bf P(X\leq t)=1-e^{-t/6}\Rightarrow P(X>t)=1-(1-e^{-t/6})=e^{-t/6}$

and we must find a t such that P(x>t)=0.5.

$\bf P(X>t)=0.5\Rightarrow e^{-t/6}=0.5\Rightarrow -t/6=ln(0.5)\Rightarrow t=-6ln(0.5)=4.16\;years$

b)

$\bf \lambda =40$

$\bf P(X>t)=0.5\Rightarrow e^{-t/40}=0.5\Rightarrow -t/40=ln(0.5)\Rightarrow t=-40ln(0.5)=27.73\;years$

c)

$\bf \lambda =344$

$\bf P(X>t)=0.5\Rightarrow e^{-t/344}=0.5\Rightarrow -t/344=ln(0.5)\Rightarrow t=-344ln(0.5)=238.44\;years$

d)

$\bf \lambda =5584$

$\bf P(X>t)=0.5\Rightarrow e^{-t/5584}=0.5\Rightarrow -t/5584=ln(0.5)\Rightarrow t=-5584ln(0.5)=3870.53\;years$

5 0

4 years ago

Hi :((( please help me solve questions 2 and 3 (the image below)

Marianna [84]

Answer:

See below

Step-by-step explanation:

Question 2:

A relation is a function if every different x-value has a y-value. Since this relation has two outputs for x = 4 (1 and 9), it is NOT a function.

Question 3:

Plug the three points listed (-1, 0, 1) into the function and get their outputs. This will be the range.

f(x) = 2x + 4

Plug in -1.

f(-1) = 2(-1) + 4

f(-1) = -2 + 4

f(-1) = 2

f(x) = 2x + 4

Plug in 0.

f(0) = 2(0) + 4

f(0) = 4

f(x) = 2x + 4

Plug in 1.

f(1) = 2(1) + 4

f(1) = 2 + 4

f(1) = 6

This makes the range {2, 4, 6}.

8 0

3 years ago