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3 years ago

13

2 is 10 times 0.02 tire of false

1 answer:

chubhunter [2.5K]3 years ago

4 0

This is false because when you multiply a number by 10, you simply move the decimal place one to the right. In this case, when we multiply 0.02 by 10, we get .2. Had we multiplied 0.02 by 100, we would have gotten 2.

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A circle has a diameter of 10 cm. What is the best approximation of its area? Use 3.14 to approximate for π .

Tasya [4]

Diameter = 10cm
radius =10/2 = 5 cm

Area = πr² = 3.14 * 5² = 78.5cm²

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3 years ago

Read 2 more answers

The answers for this problem I need to do.

lilavasa [31]

Answer:

<em>Answer is</em><em> </em><em>2x</em><em>=</em><em>1</em><em>2</em><em>8</em><em> </em><em>and</em><em> </em><em>x</em><em>=</em><em>64</em>

Step-by-step explanation:

<em> $two \: parallel \: lines \: cut \: by \: a \: transversal \: line \\ made \: alternate \: angles \: are \: equal \: \\ therefore \: 2x = 128 \\ x = \frac{128}{2} \\ x = 64$ </em>

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<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

3 0

3 years ago

Which numbers are written in scientific notation check all that apply

garik1379 [7]

Answer:

2.7 x 10⁻³

6.1 x 10⁵

9.582 x 10⁶

Step-by-step explanation:

The first digits part should be more or equal 1 and less 10

8 0

3 years ago

Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times

Elina [12.6K]

Answer:

The probability is $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is k = 1

So From the question we are told that

The number produced by factory A is 4k = 4

The number produced by factory B is 7k = 7

The probability of defective computers from A is $P(A) = 0.04$

The probability of defective computers from B is $P(B) = 0.02$

The probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is

$P(a) = 4 * P(A)$

substituting values

$P(a) = 4 * 0.04$

$P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is

$P(b) = 7 * P(B)$

substituting values

$P(b) = 7 * 0.02$

$P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is

$P(c) = 1 * P(C)$

substituting values

$P(c) = 1 * 0.03$

$P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is

$P(t) = P(a) + P(b) + P(c)$

substituting values

$P(t) = 0.16 + 0.14 + 0.03$

$P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

$P(A') = \frac{P(a)}{P(t)}$

$P(A') = \frac{ 0.16}{0.33}$

$P(A') = 0.485$

3 0

3 years ago

Log32- log 32-log 4. simplify

Rufina [12.5K]

Answer:

huh

Step-by-step explanation:

...

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.

.

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.............

4 0

2 years ago