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2 years ago

What is the y-intercept of a line perpendicular to y = x + 6 that passes through the point (4 ,-3)?

Mathematics

1 answer:

Reika [66]2 years ago

3 0

Answer:

I believe it to be C

Step-by-step explanation:

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Need help asap !!!!!

mixer [17]

First distribute the -3 to the 8 and -11y
so - 13y -24 +33y
combine like terms
so -13y and 33y together
that will be 20y -24

Hope this helps ;)

8 0

3 years ago

Read 2 more answers

The total cost of attending an AHS football game can be presented by the equation C= 5t, where C is the total cost and t is the

Nitella [24]

Seven tickets, you only need to take away the "C" and substitute for $35.00 and solve for "t"

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=11%3Dx%2B1" id="TexFormula1" title="11=x+1" alt="11=x+1" align="absmiddle" class="latex-formul

svp [43]

Answer:

x = 11-1 = 10

Step-by-step explanation:

pls mark as brainliest

7 0

3 years ago

Read 2 more answers

Making and Interpreting Correlations: Mastery Test

garri49 [273]

Answer:

Graph D > Graph A > Graph C > Graph B

Step-by-step explanation:

The closer the data points are to each other along the line of best fit, the greater the value of their correlation coefficient, and vice versa.

Therefore, the graphs can be arranged in descending other (from the highest to the lowest) based on the values of their correlation coefficients as follows:

Graph D > Graph A > Graph C > Graph B

8 0

3 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0

3 years ago