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3 years ago

2407 - -8 how do you do this type of problem

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer: 2415

Step-by-step explanation:

2407 - -8

2407- -8 is the same as 2407 take away -8.

When you take away negative numbers, the answer becomes more positive

Hence 2407- -8 is the same as make 2407 more positive by 8

That means 2407- -8 is the same as 2407+8

2407+8=2415

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you go shopping and buy x shirts for $12 and y jeans for $28 the total spent is $84 write an equation in standard form that repr

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Answer:

84= 12x + 28y

Step-by-step explanation:

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3 years ago

An expedition leader estimates that a group of hikers can carry less than 550 pounds of food and equipment.

PSYCHO15rus [73]

The number of people in the expedition is the 14.

According to the statement

we have to find that the number of people in the expedition.

So, For this purpose, we know that the

A linear equation is an algebraic equation of the form y=mx+b. involving only a constant and a first-order (linear) term, where m is the slope and b is the y-intercept.

According to the statement

The group must carry a total of 336 pounds of equipment as well as each person must carry 25 pounds of food.

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2 years ago

If a publisher of nontechnical books takes great pains to ensure that itsbooks are free of typos, so that the probability of any

laiz [17]

Answer:

(a) The probability of exactly 1 page has error is 0.271.

(b) The probability that there are at most 3 pages has error is 0.857.

Step-by-step explanation:

Let X = number of typos.

The probability of a typo is, P (X) = p = 0.005.

The number of pages in the novel is, n = 400.

The random variable X follows a Binomial distribution with parameter n and p.

But as the probability is very small and the sample size is too large we can use Poisson distribution to approximate the binomial distribution.

This distribution has parameter, $\lambda=np=400\times0.005=2$ .

The probability mass function of the Poisson distribution is:

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

(a)

Compute the probability of exactly 1 page has error as follows:

$P(X=1)=\frac{e^{-2}2^{1}}{1!} =\frac{0.13534\times2}{1} =0.27068\approx0.271$

Thus, the probability of exactly 1 page has error is 0.271.

(b)

Compute the probability that there are at most 3 pages has error as follows:

P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

$=\frac{e^{-2}2^{0}}{0!}+\frac{e^{-2}2^{1}}{1!}+\frac{e^{-2}2^{2}}{2!}+\frac{e^{-2}2^{3}}{3!}\\=0.13534+0.27067+0.27067+0.18045\\=0.85713\\\approx0.857$