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3 years ago

Solve the system by substitution. 5x+2y= 5 y=-2x + 3

1 answer:

3 0

5x+2(-2x+3)=5
5x-4x+6=5
x=-1

Then plug back into either to solve for y.

y=-2(-1)+3
y=2+3
y=5

Substitution means you just replace one variable with the other equation so only one variable is showing. That’s how you solve for the first variable.

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Connie's dog fido weighs 35 pounds. Her vet placed fido on a diet. What inequality can you write to find the average number of p

lawyer [7]

Fido's dog must lose 1.17 pounds each month to reach healthier weight.

Step-by-step explanation:

Given,

Weight of fido's dog = 35 pounds

Time period = 6 months

Goal weight to reach = 28 or less pounds

Let,

m be the weight to be lose each month.

Weight of dog - Time period * Weight to lose each month ≤ Goal weight

$35-6m\leq 28\\-6m\leq 28-35\\-6m\leq -7$

Dividing both sides by -6

$\frac{-6m}{-6}\leq \frac{-7}{-6}\\\\m\leq 1.17$

Fido's dog must lose 1.17 pounds each month to reach healthier weight.

Keywords: division, subtraction

Learn more about division at:

#LearnwithBrainly

7 0

4 years ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

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

$Z = \frac{5.8-6}{0.2}$