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3 years ago

Usethe distributive property to express28+42

Mathematics

2 answers:

nikdorinn [45]3 years ago

6 0

Answer: 7 (4 + 6)

Step-by-step explanation:

Both number are divisible by 7, so we divide them both by 7.

28/7 = 4 and 42/7 =6 . The answer is written as 7 (4 + 6)

tankabanditka [31]3 years ago

4 0

Answer:

To express the sum of two numbers using distributive property, we factor out the highest common factor (HCF) of the two numbers (i.e. the greatest number that can divide the two numbers without remainder)

28 = 2 x 2 x 7

42 = 2 x 3 x 7

The HCF of 28 and 42 is given by 2 x 7 = 14

Therefore, we express 28 + 42 using distributive property thus: 14(2 + 3)

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Answer:

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rectanlge-circle=scarf

rectanlge=28 by 15=28*15=420

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Step-by-step explanation:

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2 years ago

Six customers enter a three-floor restaurant. Each customer decides on which floor to have dinner. Assume that the decisions of

Firdavs [7]

<h2>Answer:</h2>

The probability that exactly one customer dines on the first floor is:

0.26337

<h2>Step-by-step explanation:</h2>

We need to use the binomial theorem to find the probability.

The probability of k success in n experiments is given by:

$P(X=k)=n_C_k\cdot p^k\cdot (1-p)^{n-k}$

where p is the probability of success.

Here p=1/3

( It represents the probability of choosing first floor)

k=1 ( since only one customer has to chose first floor)

n=6 since there are a total of 6 customers.

This means that:

$P(X=1)=6_C_1\times (\dfrac{1}{3})^1\times (1-\dfrac{1}{3})^{6-1}\\\\\\P(X=1)=6\times (\dfrac{1}{3})\times (\dfrac{2}{3})^5\\\\\\P(X=1)=0.26337$

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Step-by-step explanation:

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2 years ago

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a random sample of 4 claims are selected from a lot of 12 that has 3 nonconforming units. using the hypergeometric distribution

Sloan [31]

Answer:

The probability that the sample will contain exactly 0 nonconforming units is P=0.25.

The probability that the sample will contain exactly 1 nonconforming units is P=0.51.

Step-by-step explanation:

We have a sample of size n=4, taken out of a lot of N=12 units, where K=3 are non-conforming units.

We can write the probability mass function as:

$P(x=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

where k is the number of non-conforming units on the sample of n=4.

We can calculate the probability of getting no non-conforming units (k=0) as:

$P(x=0)=\frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}}=\frac{1*126}{495}=\frac{126}{495} = 0.25$

We can calculate the probability of getting one non-conforming units (k=1) as:

$P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}=\frac{3*84}{495}=\frac{252}{495} = 0.51$

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3 years ago