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2 years ago

How many pairs of whole numbers that have a sum of 12

1 answer:

6 0

<span>Total numbers of whole number that has a sum of 12. Let’s start counting
1. 11 + 1 = 12
2. 10 + 2 = 12
3. 9 + 3 = 12
4. 8 + 4 = 12
5. 7 + 5 = 12
6. 6 + 6 = 12
7. 12 + 0= 12

Therefore, there are 7 pairs of whole numbers being added that have a sum of 12.

</span>

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A random survey of the quality of city park is taken.sixty-five percent of the people surveyed were pleased with the condition o

Maurinko [17]

Answer:

280 people

Step-by-step explanation:

Given that 65% of people surveyed were pleased with condition of the park.

Given that those pleased were 182 people, then finding the total number of people surveyed will be;

65%= 182 people

100%= ?--------------------cross multiply

$\frac{65}{100} *x=182\\\\\\65x=18200\\\\\\x=\frac{18200}{65} \\\\\\x=280$

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Sue Jones is insured for bodily injury in the amount of 10/20. She is at fault in an accident in which Albert Smith is injured.

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Solution:

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2 years ago

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2 years ago

How many solutions does the following equation have? 60z+50-97z=-37z+4960z+50−97z=−37z+49

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2 years ago

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

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

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

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2 years ago