Find the arc length

A gardener sold 72 apples from the crop which was 12 % of the total

Westkost [7]

Divide the number sold by the percentage:

72 / 0.12 = 600

Total crop was 600 apples.

7 0

3 years ago

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What is the mixed number for 56/10

andre [41]

Well, 10 can go into 56, 5 times, so your whole number would be 5.
Then 50 (where did i get 50 from? 5 times the ten is 50) minues 56 is 6. So it would be :
5 and 6/10 (your denominator stays the same)
But it needs o be simplafied to its lowest terms. They are both divisable by 2, SOO 6 divided by 2 is 3, and 10 divided by 2 is 5 So, your new fraction is:
5 and 3/5 in which it cant be simplafied anymore.
<em>~ 5 3/5 </em><em>is your answer :)</em>

4 0

3 years ago

Which of the following is positive? -(-31), -(31), -31, 0-31

BartSMP [9]

It would be -(-31) because a negative and negative is a positive

3 0

3 years ago

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How many total possible outcomes are there on a standard 6 sided die?

Alenkasestr [34]

Answer:

6 (may change if multiple number cubes are used; 2 dice = 6^2, 3=6^3, so on)

7 0

2 years ago

A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig

jenyasd209 [6]

Answer:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$

The p value for this case would be given by:

$p_v =P(z$

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

$\bar X=19.87$ represent the sample mean

$\sigma=0.4$ represent the population deviation

$n=25$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$

The p value for this case would be given by:

$p_v =P(z$

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

7 0

3 years ago