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3 years ago

Is this correct? ILL MARK YOU BRAINIEST

Mathematics

2 answers:

Vsevolod [243]3 years ago

6 0

Answer:

yes

Step-by-step explanation:

EleoNora [17]3 years ago

5 0

This is correct.

Since every operation in this is multiplication, you can remove the parentheses. This is the associative property of multiplication. Now you can group the exponents and the normal factors together.

$(1.5*2.0)*(10^{0} *10^{-5})$

$3.0*10^{-5}$

So the answer you put is correct. But, if your test is online and has a dumb grading system (like edulastic) I suggest you put 3.0 instead of 3.

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Allisa [31]

Answer:

143 over 428

Step-by-step explanation:

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3 years ago

Sam is paid time and a half for overtime, if his basic hourly rate is $12.50 and he works 8 hours overtime, how much did he earn

Vesna [10]

It would be $100 if he works for eight hour hours if it wasn’t overtime but since it’s overtime and he works half of the amount it would be 50

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3 years ago

Can someone explain how to get the answer for the following question ?

zhenek [66]

The answer to the question is 1-5

3 0

3 years ago

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a

igor_vitrenko [27]

Answer: $P=0.975^t$

Step-by-step explanation:

We know that the general form of the exponential decay formula is

$y=A(1-r)^t$ , where y is final amount remaining after t time, A is the original amount and r is the rate of decay

Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. $P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}$

Hence, the ratio of remaining since the nuclear accident is $P=0.975^t$

3 0

3 years ago

Read 2 more answers

Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even

Allushta [10]

Answer:

$P(E|A)= \frac{10}{11}$

Step-by-step explanation:

Given

Two rolls of die

$E \to$ one of the outcomes is 6

$A \to$ atleast one is 6

Required

P(E|A)

First, list out the outcome of each

$E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

So:

$P(E|A)= \frac{n(E\ n\ A)}{n(A)}$

Where:

$E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$n(E\ n\ A) = 10$

$n(A) = 11$

So:

$P(E|A)= \frac{10}{11}$

7 0

3 years ago