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2 years ago

How to write 24.58 form? I got something different is this right or wrong.

Mathematics

2 answers:

Vika [28.1K]2 years ago

5 0

Answer:

That is correct.

Step-by-step explanation:

2 times 10=20

4 times 1=4

5 times 0.1=0.5

8 times 0.01=0.08

20+4+0.5+0.08=24.58

That is correct.

Sergeeva-Olga [200]2 years ago

3 0

Oh now I see. Yes that is correct

So this is what I got before, in the other question you asked, and the expanded form on the paper is the same thing as this. The way I wrote it is just another version, both are correct:

20.00 + 4.00 + 00.50 + 00.08

(2 x 10) + (4 x 1) + (5 x 0.1) + (8 x 0.01)

Multiply all the numbers together in each parentheses set and you'll get the same thing as the other problem I answered!

In the first set of parentheses (2 x 10) when solved is equal to 20

20.00 + 4.00 + 00.50 + 00.08

In the second set of parentheses (4 x 1) when solved is equal to 4

20.00 + 4.00 + 00.50 + 00.08

In the third set of parentheses (5 x 0.1) when solved is equal to .5

20.00 + 4.00 + 00.50 + 00.08

In the fourth set of parentheses (8 x 0.01) when solved is equal to 00.08

20.00 + 4.00 + 00.50 + 00.08

Hope this was way more helpful!

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2 years ago

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Answer:

(a) The value of x is 0.30.

(b) The probability that a reported case of rabies is not a raccoon is 0.45.

Step-by-step explanation:

Denote the events as follows:

R = reported case of rabies is a raccoon

F = reported case of rabies is a fox

B = reported case of rabies is a bat

O = reported case of rabies is a some other animal.

The data provided is:

P (R) = 0.55

P (F) = 0.11

P (B) = 0.04

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(a)

A property of a probability distribution is that the sum of all individual properties is 1.

That is, $\sum P(X)=1$

Compute the value of x as follows:

$\sum P(X)=1\\P(R)+P(F)+P(B)+P(O)=1\\0.55+0.11+0.04+x=1\\0.70+x=1\\x=1-0.70\\x=0.30$

Thus, the value of x is 0.30.

(b)

The probability of the complement of an event is the probability of its not happening.

$P(A^{c})=1-P(A)$

Compute the probability that a reported case of rabies is not a raccoon as follows:

$P(R^{c})=1-P(R)\\=1-0.55\\=0.45$

Thus, the probability that a reported case of rabies is not a raccoon is 0.45.

(c)

Compute the probability that a reported case of rabies is either a bat or a fox as follows:

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Thus, the probability that a reported case of rabies is either a bat or a fox is 0.15.

4 0

2 years ago