All categories

3 years ago

I need help with elimination with substitution. The equations are 3x+4y=19 and 3x+6y=33

1 answer:

4 0

For elimination, multiply one whole equation by negative one (-1), then add or subtract according to your signs. After that, it will be a one-step equation.

3x + 4y = 19 3x + 4y = 19 3x + 4y = 19 -2y = -14 y = 7
3x + 6y = 33 -1 (3x + 6y = 33) -3x - 6y = -33 -14 / -2

Then you would go back and substitute the value of (y) back into either equation and then solve for the remaining variable (x). Finally, use both values to make an ordered pair.

3x + 4y = 19 3x = -9 x = -3 (-3 , 7)
3x + 4(7) = 19 (-9 / 3)
3x + 28 = 19

Good Luck

You might be interested in

Which shows the best path to conert 100 liters to pints

lions [1.4K]

1 liter equals to 2.11338 pints

When converting 100 liters to pints you would do 100 * 2.11338, or move the decimal place two places to the right, so you have 211.338 pints.

8 0

3 years ago

Read 2 more answers

HEEEELPPPPP Which is the graph of the equation

krek1111 [17]

the answer is........ C

8 0

3 years ago

PLEASE HELP URGENT!!!!!!!

MrMuchimi

Answer:

Choice B is correct

Step-by-step explanation:

4 0

3 years ago

What store is offering the best sale <br> 8 for $6<br> 10 for $10<br> 2 for $3<br> 0.80 cents each

Alex

Answer:

2 for 3

Step-by-step explanation:

8÷6=$1.33

10÷10=$1.00

2÷3=$0.66

.80

2 for $3 is less than the rest therefore it is a better deal

6 0

2 years ago

In studies for a medication, 3 percent of patients gained weight as a side effect. Suppose 643 patients are randomly selected.

timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

$p = 0.03$

$q = 1 - 0.03 = 0.97$

The mean is

$\mu = np$

$\mu = 643 \times 0.03 = 19.29$

The standard deviation is

$\sigma = \sqrt{npq}$

$\sigma = \sqrt{643 \times 0.03 \times 0.97}$

$\sigma =4.33$

We want to find the probability that exactly 24 patients will gain weight as side effect.

P(X=24)

We apply the Continuity Correction Factor(CCF)

P(24-0.5<X<24+0.5)=P(23.5<X<24.5)

We convert to z-scores.

$P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 1.20) \\ = 0.051$

Part b) We want to find the probability that 24 or fewer patients will gain weight as a side effect.

P(X≤24)

We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

$P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849$

Part c)

We want to find the probability that

11 or more patients will gain weight as a side effect.

P(X≥11)

Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

$P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)$

$= 0.9788$

Part d)

We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

P(23.5≤X≤28.5)

Convert to z-scores:

$P(23.5 \: < \: X \: < \: 28.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{28.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 2.13) \\ = 0.1494$

3 0

3 years ago