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3 years ago

The diagram shows a rectangle.

Mathematics

2 answers:

8090 [49]3 years ago

4 0

Answer:

To solve this we calcuclate the total are then we subtract the 4 rectangles area

total area = length × width =10.5 × 10.5 = 110.25 cm²

the rectangles drawn accurately = 2(.5 × 7)

7 cm²

the rectangles drawn not accurately 2(.5 × 10.5)

= 10.5 cm²

so the shaded area = 110.25 - 7 - 10.5 = 92.75 cm²

Taya2010 [7]3 years ago

4 0

Answer:

42 cm²

Step-by-step explanation:

We can see that the length of the blue rectangle is the same as the length of the smaller rectangle = 7 cm.

To calculate the width of the blue rectangle, we take the length of the smaller rectangle (7 cm) and substract two of the width (0.5 cm) of the smaller rectangle:

7 - (2 x 0.5) = 6 cm

Now we have the length and width of the blue rectangle.

Are of blue rectangle = l x w = 7 x 6 = 42 cm²

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WITCHER [35]

Answer:

C) 2+7i is the correct answer.

Step-by-step explanation:

simplify by combining the real and imaginary parts of each expression. in this question, the variable i represents an imaginary number.

8 0

3 years ago

Which statement best describes the area of Triangle ABC shown below?

PolarNik [594]

Answer:

it is half the area of a square of side length 8

Step-by-step explanation:

area of a square = l × b

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= 64 cm^2

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= 8 × 8 ÷ 2

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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

3 0

3 years ago

A graph of a system of two linear equations is shown. Write the system of linear equations represented by the graph

DiKsa [7]

Line1: y = -x + 2
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2 years ago

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frutty [35]

Answer:

Step-by-step explanation:

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3 years ago