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3 years ago

Test 69,902 for divisibility by 2, 3, 5, 9, or 10.

Mathematics

1 answer:

Keith_Richards [23]3 years ago

5 0

Its only divisible by 2.

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At the snack bar, a hot dog costs $1.50 and a slice of pizza costs $2. What percent of the cost of a slice of pizza is a cost of

ZanzabumX [31]

That percent is 133
There are several answers for this
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5 0

3 years ago

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Use a quadratic equation to find two numbers whose sum is 5 and whose product is - 24.

prisoha [69]

Hello
if m s : sum and p : product
the quadratic equation is : x²-sx+p=0
s = 5 p=-24
x²-5x-24 = 0

8 0

3 years ago

Caiden and Josh are practicing at a batting cage. Josh has hit nine more than one-half the number of balls that Caiden has. If t

OverLord2011 [107]

Answer:

The Answer is: Josh hit 47 balls.

Step-by-step explanation:

Let j = Josh number of hits and c = Caiden number of hits:

j = c/2 + 9

Total:

j + c = 123.

Substitute:

(c/2 + 9) + c = 123

c/2 + c = 123 - 9

c/2 + c = 114

1/2c + 2/2c = 114

3/2c = 144

c = 144 (2/3) = 76, Number of hits for Caiden

Solve for j:

j = 76/2 + 9

j = 38 + 9 = 47, number of hits for Josh

Proof:

j + c = 123

47 + 76 = 123

123 = 123

Hope this helps! Have an Awesome Day!! :-)

6 0

3 years ago

Graph your linear equation. y=x÷3

mojhsa [17]

Here ya go
(correct me if wrong)

7 0

2 years ago

In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that

attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let X = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0

3 years ago