All categories

3 years ago

The graph of which of the following equations (0,-3)

Mathematics

1 answer:

Bad White [126]3 years ago

7 0

A Because y=mx+b So if you graphed it then it would be A

You might be interested in

A corn vendor at a farmer’s market was selling a bag of 8 ears of corn for $2.56. Another vendor was selling a bag of 12 for $4.

Sliva [168]

Answer:

The bag of 8 ears of corn

Step-by-step explanation:

2.56/8 = .32 per ear of corn

4.32/12 = .36 per ear of corn

8 0

3 years ago

Twelve pounds of beans are distributed equally into 8 bags to give out at the food bank. How many pounds of beans are in each ba

BaLLatris [955]

There are 1.5 beans in 1 bag

5 0

3 years ago

A receipe calls for 3/8 cup of chocolate chips and 1 1/2 cups of walnuts. Tameka wants to double the recipe. How many cups of wa

kirill115 [55]

Answer:

3 cups

Step-by-step explanation:

1 1/2 + 1 1/2 = 3

8 0

3 years ago

Who knows about ratios

Kaylis [27]

I can help you. Just post the question

5 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