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3 years ago

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How do I write the number 17,497,403 well i know how to do it inside standard but my equation says “Write the number Seventeen m

illion, four hundred ninety-seven thousand, four hundred three in standard form and expanded form using multiplication. And I don’t know how to do that

1 answer:

MrRa [10]3 years ago

7 0

Step-by-step explanation:

Standard form is of course 17,497,403.

Expanded form is:

10,000,000 + 7,000,000 + 400,000 + 90,000 + 7,000 + 400 + 3

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In a certain Algebra 2 class of 29 students, 7 of them play basketball and 24 of them play baseball. There are 3 students who pl

antiseptic1488 [7]

Answer:

$\frac{5}{29}$

Step-by-step explanation:

Let $n(A)$ represent students playing basketball, $n(B)$ represent students playing baseball.

Then, $n(A)=7$ , $n(B)=24$

Let $n(S)$ be the total number of students. So, $n(S)=29$ .

Now,

$P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}$

$P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}$

3 students play neither of the sport. So, students playing either of the two sports is given as:

$n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26$

∴ $P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}$

From the probability addition theorem,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Where, $P(A\cap B)$ is the probability that a student chosen randomly from the class plays both basketball and baseball.

Plug in all the values and solve for $P(A\cap B)$ . This gives,

$\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}$

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{5}{29}$

6 0

3 years ago

Suppose a shipment of 120 electronic components contains 3 defective components. to determine whether the shipment should be ac

Likurg_2 [28]

For this problem, the most accurate is to use combinations

Because the order in which it was selected in the components does not matter to us, we use combinations

Then the combinations are $nC_r = \frac{ n! }{r! (n-r)!}$

n represents the amount of things you can choose and choose r from them

You need the probability that the 3 selected components at least one are defective.

That is the same as:

(1 - probability that no component of the selection is defective).

The probability that none of the 3 selected components are defective is:

$P = \frac{_{117}C_3}{_{120}C_3}$

Where $_{117}C_3$ is the number of ways to select 3 non-defective components from 117 non-defective components and $_{120}C_3$ is the number of ways to select 3 components from 120.

$_{117}C_3 = 260130$

$_{120}C_3 = 280840$

So:

$P = \frac{260130}{280840} = 0.927$

Finally, the probability that at least one of the selected components is defective is:

$P = 1-0.927 = 0.0737$

P = 7.4%

3 0

4 years ago

You have purchased a property that you intend to rent out to people. The mortgage payment is $765/month and the property taxes a

Delicious77 [7]

Monthly mortgage payment multiplied by 12 months in a year.
765*12=9,180.
Add yearly mortgage plus property taxes:
9,180+4,056=13,236
Divide the total by 12 months in a year.
13,236/12=1,103
You should charge $1,103 a month to come out even.

You can also take the property taxes and divide them by 12 (months in a year) 4056/12=338. Then add the monthly mortgage ($765)
765+338= $1,103.

Hope this helps :)

7 0

3 years ago

Please Help! GEOMETRY!

lara31 [8.8K]

I think the answer should be A n goodluck

3 0

4 years ago

Read 2 more answers

I need to know how to do this

kiruha [24]

1) 2
2) 3
3) 4
4) 7
Hope this helps ^-^

5 0

3 years ago