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2 years ago

(a-b+c)-(c+b) I literally tried so many times i dont get it

Mathematics

2 answers:

Sauron [17]2 years ago

5 0

Answer:

a - 2b

Step-by-step explanation:

❖ Step 1: Distributive negative sign.

$(a-b+c)+(-1)(c)+(-1)(b)$
$a-b+c-c-b$

❖ Step 2: Combine like terms.

$(a)+(-b-b)+(c-c)$
$a-2b$

Therefore, the answer is a - 2b.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

ollegr [7]2 years ago

4 0

Answer:

a-2b

Step-by-step explanation:

(a-b+c)-(c+b)

Distribute the minus sign to the terms in the second parenthesis

a-b+c-c-b

Combine like terms

a-b+c-c-b

a-b-b+c-c

Cancel like terms

a-2b

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Answer:

£12.60

Step-by-step explanation:

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3 years ago

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3 years ago

Write an equation to solve for x 3x + 21 6x-60 will mark brainiest

Maru [420]

Answer:

Step-by-step explanation:

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3 years ago

The weights of a certain dog breed are approximately normally distributed with a mean of 49 pounds, and a standard deviation of

AlekseyPX

Answer:

a. 74.86%

b. 50%

c. 50%

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of 49 pounds, and a standard deviation of 6 pounds.

This means that $\mu = 49, \sigma = 6$

a. Find the percentage of dogs of this breed that weigh less than 53 pounds.

The proportion is the p-value of Z when X = 53. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{53 - 49}{6}$

$Z = 0.67$

$Z = 0.67$ has a p-value of 0.7486.

0.7486*100% = 74.86%, which is percentage of dogs of this breed that weigh less than 53 pounds.

b. Find the percentage of dogs of this breed that weigh less than 49 pounds.

p-value of Z when X = 49, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49 - 49}{6}$

$Z = 0$

$Z = 0$ has a p-value of 0.5

0.5 = 50% of dogs of this breed that weigh less than 49 pounds.

c. Find the percentage of dogs of this breed that weigh more than 49 pounds.

1 subtracted by the p-value of Z when X = 49, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49 - 49}{6}$

$Z = 0$

$Z = 0$ has a p-value of 0.5.

1 - 0.5 = 0.5 = 50% of dogs of this breed that weigh more than 49 pounds.

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