3X minus 2Y is equal to or greater than or equal to four

How to solve using substitution y=3/4X+5 4x-3y=-1

Pachacha [2.7K]

Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as y=(3/4)X+5

Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:

4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)

Let 's now solve the system

4y=3x + 20
4x-3y=-1

We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:

12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)

Then we have

12y = 9x + 60
-12y =-16x - 4

If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8

We were given equations
y=3/4X+5
4x-3y=-1

We can subst. x=8 into either of these eqn's to find y. Let's try the first one:

y = (3/4)(8)+5 = 6+5=11

Then x=8 and y=11.

You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?

5 0

3 years ago

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5

lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 0.499, \sigma = 0.002$

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

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

$Z = \frac{0.504 - 0.494}{0.002}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

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

$Z = \frac{0.496 - 0.494}{0.002}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

4 0

3 years ago

Which of the following theorems verifies that ABC ~ STU?

Lunna [17]

Answer:

B. AA

Step-by-step explanation:

The diagram given shows that two angles in ∆ABC are congruent to two corresponding angles in ∆STU.

Invariably, the third unknown angle of both triangles would also be equal going by the third angle theorem.

Thus, based on the AA Similarity Theorem which says that two triangles are similar to each other if two corresponding angles of one is congruent to two angles in the other, ∆ABC ~ ∆STU.

4 0

2 years ago

There are two functions that can be used to describe the percent of people 25 years and older who have completed at least high s

Korvikt [17]

Answer:

$g(15)=93.8\%$

see the explanation

Step-by-step explanation:

Let

x ----> the number of years since 2008,

g(x) ----> the percent of people 25 years and older who have completed at least high school

we know that

$g(x)=0.008x^2+0.42x+87.5$

so

For x=15 years since 2008

substitute

$g(15)=0.008(15)^2+0.42(15)+87.5=93.8\%$

That means

In the year 2023 (2008+15) the percent of people 25 and older completing at least high school will be 93.8%

5 0

2 years ago

The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in

Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 64 inches

$\sigma$ = standard deviation = 4 inches

n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{63-64}{\frac{4}{\sqrt{4} } }$ ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

= 1 - 0.69146 = 0.30854

Hence, it is 30.85% likely that average height would be shorter than 63 inches.

7 0

3 years ago