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3 years ago

10

Volunteers were 3/4 finished with the walkathon when they reached the 4 1/2 kilometer mark. How many kilometers was the walkatho

n

1 answer:

den301095 [7]3 years ago

5 0

5 I think, cause if your 3/4 ways done you need to walk 1/2 more kilometers to be finished... So if you add 1/2 to 41/2 it equals 5

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If x=-6, which inequality is true ??<br>

Gelneren [198K]

<em>Answer:</em>

The inequality that is true is -5 - 3x > 10.

<em>Step-by-step explanation:</em>

if x = -6

we can subtitution x to every inequality, so :

-5 - 3x > 10

-5 - 3(-6) > 10

-5 - (-18) > 10

-5 + 18 > 10

13 > 10 [true]

-3 - 5x < -14

-3 - 5(-6) < -14

-3 - (-30) < -14

-3 + 30 < -14

27 < -14 [false]

2 - x < -3

2 - (-6) < -3

2 + 6 < -3

8 < -3 [false]

So if x=-6, inequality true is -5 - 3x > 10.

<em>If this was helpful, please mark brainliest. Have a beautiful day. </em>

4 0

3 years ago

Help me someone!!!!!!!!!!!!!!!!!!!!!!!

Julli [10]

Substitute answers in, a = 3 and b = 2.

(3^2 - 2^2)/(3+2)
= (9-4)/5
= 5/5
= 1

7 0

3 years ago

Two numbers are in the ratio 7:9. The smaller number is 42. what is the number?

ddd [48]

Answer:

7/9 = 42/n.

7n = 378,

n = 54.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

8 0

3 years ago

During a sale, for each shirt that Mark purchased at the regular price, he also purchased a shirt at half the regular price. How

Arisa [49]

I think the answer is b hope i helped

4 0

4 years ago