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3 years ago

Write the numbers in scientific notation.

Mathematics

1 answer:

Rasek [7]3 years ago

3 0

Answer:

8)5.1*10⁶

9)6.98*10 to the power of -6

10) 3.000052*10⁰

11)0.006548

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Mr Fadzil used 1 2/5m of cloth to make 2 identical phone cases and a cushion cover. He used 3/5

loris [4]

Answer:

1/5

Step-by-step explanation:

2/5 + 3/5

(2+ 3)/5

2/5 + (2 + 3)/5 = (4 + 3)/5

(4+ 3)/5 = 7/5

4+ 3 = 7

4 = 4

c = 1/5

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3 years ago

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Ksju [112]

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4 years ago

What is the quotient of 36,772 and 58?

kozerog [31]

Answer:

634.

Step-by-step explanation:

Quotient meaning:

To divide 2 numbers.

Equation:

36,772 ÷ 58

Solve:

36,772 ÷ 58
= 367 - 348 = 19
= 197 - 174 = 23
= 232 - 232
= 0

Multiples:

6, 3, and 4.

Answer:

634.

4 0

3 years ago

Read 2 more answers

Rachel spent $70 more than Samuel. If Samuel spent $449, then how much did Rachel spend?

tatiyna

Answer: Rachel spent $519

Step-by-step explanation: Add 70 to 499 and you get 519

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3 years ago

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All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (A

Semenov [28]

Answer:

a ) the probability of using steroids and having a negative test is 0.5%

b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following

$P(A\cupB) = P(A)+P(B)-P(A\cap B) = P(A) +P(B)-P(A|B)P(B) = 6.4\%+5\%-90\%\cdot 5\%=6.9\%$

4 0

4 years ago