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White raven [17]
2 years ago
12

Joshua babysits for his sister after school. He is paid

Mathematics
2 answers:
Yakvenalex [24]2 years ago
7 0

Answer:

9 hrs

Step-by-step explanation:

Joshua has already saved $43.50 so you subtract $43.50 from 70 which leaves you with $26.50. Joshua still needs $26.50.

$26.50 divided by $3 per hr is 8.833 repeating. Obviously you can’t work 8.833 hrs so you round it up to 9. So, Joshua has to work for at least 9 hrs more.

Hope this helps :)

Darina [25.2K]2 years ago
4 0
The answer is 14.50. Hope that helped
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Kruka [31]

Answer:

O=30° or O=29°

Step By Step Explanation:

∠α =arcsin( a/c )

= arcsin(4/8)

= arcsin(0.5)

= 0.5 rad = 30° = 29°

I Hope This Helps. I Gave you 2 answers just in case 1 is Wrong. And I Hope you Understand This Calculation, I Used Different Variables and the arcsin thing is something my teacher told my class to use, and I Have no Idea what it means because my teacher never explained what it meant, Sadly.

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Answer:

Step-by-step explanation:

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2 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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3 years ago
Given side AB in ΔABC has length x, what is the length of side AC?
gladu [14]

Answer:

2x

Step-by-step explanation:

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2 years ago
A sport store ordered 22 boxes of golf balls. Each box has 47 golf balls. Which is the best estimate of the total number of golf
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About 1,000! Hope this helps
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3 years ago
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