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3 years ago

Help me out please simple 8th grade math.

Mathematics

1 answer:

galben [10]3 years ago

6 0

X-3
The blank is minus 3

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If the y-intercept of the function y=4x-7 were increased by 10 what would be the equation of the new function

Sonja [21]

The equation would be:
$y = 4x + 3$

Hope this helps.

8 0

4 years ago

Read 2 more answers

0.5x -2 = 1.5x + 0.15

frutty [35]

Answer: x=−2.15

Step-by-step explanation : Alright Let's solve your equation step-by-step.

0.5x−2=1.5x+0.15

Step 1: Subtract 1.5x from both sides.

0.5x−2−1.5x=1.5x+0.15−1.5x

−x−2=0.15

Step 2: Add 2 to both sides.

−x−2+2=0.15+2

−x=2.15

Step 3: Divide both sides by -1.

−x

−1

2.15

−1

x=−2.15<em> hope this helped!- Alex</em><u> </u>

8 0

4 years ago

Whats the value of the 2 in 0.812 and write the underline digit as a fraction.

Snowcat [4.5K]

Answer:

the value of 2 in 0.812 is 2 thousandths/0.002

0.002 = 2/1000

0.812 = 812/1000

3 0

3 years ago

What is the answer to X-9=8(2x+3)-18

Mariulka [41]

X-9=8(2x+3)-18
x-9=16x+24-18
-9=15x+6
-15=15x
x=-1

7 0

3 years ago

Read 2 more answers

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Portland. The stu

iVinArrow [24]

Answer:

$-0.061 < P_1 -P_2< 0.025$

Step-by-step explanation:

Give data:

$n_1 = 350$

$n_2 =250$

$x_1 = 23$

$x_2 = 21$

$\hat{P} 1 = \frac{x_1}{n_1} = \frac{23}{350} = 0.066$

$\hat{P} 2 = \frac{x_2}{n_2} = \frac{21}{250} = 0.084$

for 95% confidence interval

$\alpha = 1 - 0.95 = 0.05 and \alpha/2 = 0.025$

$z_{\alpha/2} = 1.96$ from standard z- table

confidence interval for P_1 and P_2 is

$\hat{P} 1 - \hat{P} 2 \pm z_{\alpja/2} \sqrt{\frac{\hat{P} 1(1-\hat{P} 1)}{n_1} +\frac{\hat{P} 2(1-\hat{P} 2)}{n_2}}$

$(0.066 - 0.084) \pm 1.96 \sqrt{\frac{0.066(1-0.066)}{350} +\frac{0.084(1-0.084)}{250}}$

$-0.018 \pm 0.043$

confidence interval is

$-0.018 - 0.043 < P_1 -P_2$

$-0.061 < P_1 -P_2< 0.025$

7 0

3 years ago