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2 years ago

Is 4/5 greater then 0.75

Mathematics

2 answers:

Alexxandr [17]2 years ago

6 0

Yes because 4/5 as a decimal is 0.80 so in other words yes it is bigger than 0.75

GalinKa [24]2 years ago

4 0

Yes 4/5 is greater than 0.75 because if you put 0.75 as a fraction that would be 75/100 then you do the same denominator and 5 can go into 100 20 times and so you do that to the numerator and you get 80/100 and 75/100 so 4/5 id grater than 0.75.

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The formula for the area A of a rectangle with length l and width w is A = l ⋅ w . Find A if: l = 23 feet and w = 15 feet The ar

dexar [7]

Answer:

345 sq. ft

Step-by-step explanation:

You know that the area is the length times the width. Simply plug in values to your equation:

A = l * w

A = 23 * 15

A = 345

The area is 345 square feet.

3 0

3 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

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

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

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

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

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2 years ago

Please answer it now in two minutes

Nikitich [7]

Answer:

3√6

Step-by-step explanation:

tan60=opp/adj

opp(d)=tan60*3√2=√3*3√2=3√6

7 0

3 years ago

The function g is defined by g(x) = x² +4. Find g(5n)

jasenka [17]

Answer:

g(5n)=5n^2+4

Step-by-step explanation:

Substitute 5n for x

5n^2+4

You can't really do much with this so that is the answer

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3 years ago

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Wittaler [7]

Answer:

yeah but it depends on what your measuring

have a good day :)

Step-by-step explanation:

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2 years ago