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2 years ago

The length of a diagonal of a cube with edge length 25cm

Mathematics

1 answer:

elena-s [515]2 years ago

6 0

Let's take any edge length l. The diagonal has length $l\sqrt3$ . In our case it's $25\sqrt3$

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9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will rema

DerKrebs [107]

Answer:

1 Shirt and 185cm remaining cloth

Step-by-step explanation:

2m and 15cm = 215cm

40m = 400cm

400/215=1.860465116

A shirt cant be in decimal so we remove the decimals, and take the as the answer. For the decimals, they are the remaining cloth.. to calculate the amount of remaining cloth we do:

1.860465116-1

=860465116

215*860465116=185cm

5 0

2 years ago

What is the median of this set of data? 1, 2, 5, 6, 9 1 5 8 9

gavmur [86]

Answer:

5.5

Step-by-step explanation:

1, 2, 5, 5, 6, 8, 9, 9

5+6=11

11 divided by 2= 5.5

4 0

3 years ago

Read 2 more answers

kathy uses a 1/2 cup of milk with every bowl of her favorite cereal if there are only 3 3/5 cups of milk left, then how many bow

Murrr4er [49]

Taking a quotient, we will see that she can make 7 bowls of cereal (and some leftover milk).

<h3>How many bowls of cereal would kathy have?</h3>

We know that for each bowl, she needs 1/2 cups of milk.

And we also know that she has a total of (3 + 3/5) cups of milk.

To know how many bowls she can make, we need to take the quotient between the total that she has and the amount that she needs for each bowl:

(3 + 3/5)/(1/2)

We can rewrite the total as:

3 + 3/5 = 15/5 + 3/5 = 18/5

Then the quotient becomes:

(18/5)/(1/2) = (18/5)*2 = 36/5 = 35/5 + 1/5 = 7 + 1/5

So she can make 7 bowls of cereal (and some leftover milk).

If you want to learn more about quotients:

brainly.com/question/629998

#SPJ1

6 0

1 year ago

Each letter in this equation represents a unique digit. If U stands for 8 and the

cricket20 [7]

Answer:

4x2=8

Step-by-step explanation:

This equation is simple and easy cause its base sick math

8 0

1 year ago

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0

2 years ago