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2 years ago

9

Diego has 4 times as manny autograhed baseballs as melanie. diego has 24 autographed balls . how many does melanie have?

1 answer:

Ainat [17]2 years ago

3 0

Answer:

Melanie has 6 autographed baseballs.

Step-by-step explanation:

Diego has 4 times as many autographed baseballs as Melanie, and Diego has 24. 24/4 = 6.

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How many times does 32 go into 82

pav-90 [236]

More specifically 32 goes into 82, 2.5625 times.

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2 years ago

Which of the following is a radical equation?<br>a.x√3=13 <br>b.x+√3=13<br>c.√x+3=13 <br>d.x+3=√13

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Answer:

this one: √x+3=13

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2 years ago

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What is the equivalent expression for 2X -4 minus X

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Answer:

Step-by-step explanation:

x ≓ -2.009940617

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3 years ago

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-5(8p - 2) - 5p = 10 + 5p

Liula [17]

Answer:

-5(8p - 2) - 5p = 10 + 5p

=> -40p + 10 - 5p = 10+ 5p

=> -40p - 5p -5p = 10 -10

=> -50p = 0

=> p = 0/(-50)

=> 0

Hope it helps.......

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2 years ago

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Suppose a certain study reported that 27.7% of high school students smoke.

Norma-Jean [14]

<u>Given</u>:

Probability of student smoke,

P = 27.7%

= 0.277

Number of students,

n = 632

$q = 1-p$

$=1-0.277$

$=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

$=16.62$

⇒ $n_q=60\times 0.723$

$=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

$= \sqrt{60\times 0.277\times 0.723}$

$=3.9664$

Now,

⇒ $P(X$

$=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

$=20.775$

$\sigma = \sqrt{n_{Pq}}$

$=\sqrt{75\times 0.277\times 0.723}$

$=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

$=1-P(X \leq 17.5)$

$=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

$=0.835$

Learn more about probability here:

brainly.com/question/9825651

4 0

3 years ago