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3 years ago

3x+4y= -24 Find the x-intercept and the y-intercept.

Mathematics

2 answers:

Lilit [14]3 years ago

7 0

Answer:

y = -3/4 - 6

Step-by-step explanation:

3x + 4y = -24

get Y to be by itself so subtract 3x on both sides

4y=-3x-24

divide by 4 on both sides

y = -3/4x -24/4

-24/4=-6

y = -3/4 - 6

Lapatulllka [165]3 years ago

4 0

Answer:

Step-by-step explanation:

x-intercept: let y = 0 and find x. Then 3x = -24, and x = -8. The x-intercept is (-8, 0).

y-intercept: Let x = 0 and find y. 4y = -24 yields y = -6. The y-intercept is (0, -6).

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ss7ja [257]

ANSWER

10 weeks, $310

EXPLANATION

We have a system of linear equations as both equations need to be satisfied by the same ordered pair (x,y):

y = 25x + 60

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We can solve this by substitution. Since y is equal to 25x+60, we can substitute in 25x + 60 for y in the equation y=30x+10

y = 30x + 10

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Subtract 10 from both sides and then divide both sides by 5

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x represents time in weeks, so 10 weeks.

Find y, the amount of money, by substituting x=10 into either equation.

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y = 30x + 10

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3 years ago

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Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two

sveta [45]

Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

Alloy A Alloy B

Sample mean $\bar {x} _A = 49.5$ $\bar {x} _B = 45.5$

Equal standard deviation $\sigma_A = 5$ $\sigma_B= 5$

Sample size $n_A = 30$ $n_{B}= 30$

Mean of the sampling distribution is :

$\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5 \\ \\ = 4.0$

Standard deviation of sampling distribution:

$\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667} \\ \\ =1.2909$

Hypothesis testing.

Null hypothesis: $H_o: \mu_A -\mu_B = 0$

Alternative hypothesis: $H_A:\mu_A -\mu_B > 4$

The required probability is:

$P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010$

This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

Since the P-value is very small which is lower than any level of significance.

Then, we reject $H_o$ and conclude that there is enough evidence to fully support alloy A.

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Answer:

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Write a two column proof 2x+6 and 96 prove: x=45

viktelen [127]

2x+6=96
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