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3 years ago

6 ÷ 3 + 32 · 4 − 2 A. 36 B. 98 C. 22 D. 42

Mathematics

1 answer:

otez555 [7]3 years ago

4 0

Answer:128

Step-by-step

2+ 32 x 4-2

2+128-2

=128

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Please help me solve

vlabodo [156]

(2x+5)=Inscribed Angle
50 = Arc Length
Inscribed Angle=1/2Arc Length
2x+5=1/2(50)
2x+5=25
2x=20
X=10

4 0

3 years ago

I need help I don't understand this

Lady_Fox [76]

<h3>Answer :</h3>

It is clear from the diagram that,

ㄥ6 + ㄥ8 = 180

⇒ (2x - 5) + (x + 5) = 180

⇒ (2x + x) + (5 - 5) = 180

⇒ 3x + 0 = 180

⇒ x = 180/3

⇒ x = 60°

◈ From the geometry of figure,

ㄥ3 = ㄥ6

⇒ ㄥ6 = (2x - 5)

⇒ ㄥ6 = (2×60 - 5)

⇒ ㄥ6 = 120 - 5

⇒ ㄥ6 = 115°

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3 years ago

Which are equivalent expressions? Check all that are true. 11a + 12b = 23ab

mash [69]

Combine the like terms to find your answer

8 0

3 years ago

Martina is currently 14 years older than her cousin Joey. In 5 years she will be 3 times as old as Joey. How old is Martina? How

Rufina [12.5K]

Answer:

Joey is 2 years old and Martine is 16 years old.

Step-by-step explanation:

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4 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

3 0

3 years ago