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3 years ago

Hi please help me with this i’ll give brainliest if you give a correct and show your work!

Mathematics

1 answer:

Tatiana [17]3 years ago

6 0

1. (1.35)x(.42)

1.35
x. .42
________
270
5400
________
.5670

You need to have the total number of decimal places from the original two numbers. 1.35 has two decimal places and .42 has to decimal places you have a total of four decimal places, Which means 5670 is .5670

2. (.22)x(.04)

.22
x. .04
_______
88
00
_______
.0088

you need to have the total number of decimal places from the original Two numbers. So .22 is two decimal places and .04 is two decimal places, so you will need to have for decimal places. That means that the answer 88 needs to be translated into .0088

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What is the slope of the line going through the points (-3,-7.5) and (-2,-5)

lakkis [162]

Answer:

2.5

Step-by-step explanation:

y2 - y1 = top

x2 - x1 = bottom

y2 = -5

y1 = -7.5

x2 = -2

x1 = -3

-5-(-7.5) = 2.5

-2-(-3) = 1

2.5 over 1 = 2.5

7 0

3 years ago

What is x and measure b? please help:)

BARSIC [14]

Answer: x= 22.5, angle B = 67.5

Step-by-step explanation:

90+3x+x= 180

4x = 90

x=22.5

3(22.5)= 67.5

5 0

3 years ago

Read 2 more answers

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

3 0

3 years ago

Brainliest for right answer!!!<3

GenaCL600 [577]

The value of a should be 2000. Hope this helps.

6 0

3 years ago

Yati has a coin box containing some coins. 40% of the coins are 20-cent coins and the rest are 50-cent coins.

lana66690 [7]

The fraction that should be removed is 1/5.

<h3>How to calculate the fraction?</h3>

It should be noted that from the information, 40% of the coins are 20-cent coins and the rest are 50-cent coins. Those that are 50 cent will be:

= 100 - 40%

= 60%

The fraction that's should be removed will be:

= 2/5 - x = 1/5

x = 2/5 - 1/5

x = 1/5

The fraction that should be removed is 1/5.

Learn more about fractions on:

brainly.com/question/17220365

#SPJ1

7 0

2 years ago