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3 years ago

Find the area of the shape shown below. 3.5,2,2,2,2,5

Mathematics

1 answer:

natta225 [31]3 years ago

4 0

Answer:

26.75

Step-by-step explanation:

It is 26.75 because if you split the traingle in half from the bottom to the top you look to find the area of each shape, so the sqaure has a measure of 2 by 2 so that is four then the smaller triangle has a measure of 2 by 2 but the formula for triangles are length times height divided by 2 and the area for that is 2. Then the other smaller triangle is 5 * 2 which is 10 then divide it by 2 so it is 5 so all together it is 9. then the area of the biggest triangle is 15.75

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Which of the following statements is TRUE about the relationship between the numerical values?

Gwar [14]

Answer:

D is the answer

Step-by-step explanation:

-15 Is the same as 15 its just negative but they have the same distance.

7 0

3 years ago

11×11=4 22×22=16 33×33=?

Kay [80]

This is your answer <span>33 x 33 = 4^3 = 64</span>

7 0

3 years ago

Find the range of ƒ(x) = 3g – 5 for the domain {–1.5, 2, 4}

timofeeve [1]

-1.5 ------ f(x) = 3(-1.5) - 5
= -4.5 -5
= 9.5

2 -------- f(x) = 3(2) - 5
= 6 -5
= 1

4 -------- f(x) = 3(4) - 5
= 12 - 5
= 7
(9.5, 1 , 7)

6 0

3 years ago

What is the average rate of change in profit from Hour 5 to Hour 10?

snow_lady [41]

Answer:

Step-by-step explanation:

so at 5 hours he was at a profit of 900 and 5 hours later he is at a profit 1500 then you can sub tract 900 from 1500 which gives you 600

8 0

3 years ago

A recipe submitted to a magazine by one of its subscribers’ states that the mean baking time for a cheesecake is 55 minutes. A t

EastWind [94]

Answer:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X-i -\bar x)^2}{n-1}}$

$\bar X=59.4$ represent the sample mean

$s=3.239$ represent the sample standard deviation

$n=10$ sample size

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:

Null hypothesis: $\mu \leq 55$

Alternative hypothesis: $\mu > 55$

Replacing the info given we got:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing the info given we got:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

8 0

3 years ago