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3 years ago

Give an example of an addition problem in which you would and would not group the addends differently to add

1 answer:

8 0

<span>Changing the grouping of the addends should not change the sum, according to the associative property of addition. You might group them differently with (50 + 3) + 47, so that you have 50 + (3 + 47). You might not regroup them with (16 + 4) + 5 rather than 16 + (4 + 5).</span>

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Find the surface area of the prism

AfilCa [17]

Answer:

156 $in^2$

Step-by-step explanation:

find the areas of all the faces and add them up

area of a rectangle: $bh$

area of a trapezoid: $\frac{1}{2} (b_1 + b_2)(h)$

$(5*3)+(6*3)+(5*3)+(3*12)+2(\frac{1}{2}*(6+12)*4) \\=15+18+15+36+2(\frac{1}{2}*18*4)\\=84+2(36)\\=156$

the area of the trapezoid is multiplied by 2 because there's a top and bottom

4 0

3 years ago

A unicycle wheel has a diameter of 12 inches. How many inches will the unicycle

saw5 [17]

Answer:The unicycle would travel 226.29 inches in 6 revolutions.

Step-by-step explanation:

8 0

3 years ago

Triangle ABC has vertices A(–1, 0), B(4, 0), and C(2, 6). If ΔABC is translated using the function rule (x, y) → (x – 6, y – 5)

Elina [12.6K]

Answer:

b. A''(7, 5), B''(2, 5), C''(4, –1)

Step-by-step explanation:

Given vertices of ΔABC:

A = (-1, 0)
B = (4, 0)
C = (2, 6)

Translation <u>mapping rule</u>:

(x, y) → (x – 6, y – 5)

Therefore:

A' = (-1 - 6, 0 - 5) = (-7, -5)
B' = (4 - 6, 0 - 5) = (-2, -5)
C' = (2 - 6, 6 - 5) = (-4, 1)

Rotation of 180° clockwise rule:

(x, y) → (-x, -y)

Therefore:

A'' = (-7, -5) = (7, 5)
B'' = (-2, -5) = (2, 5)
C'' = (-4, 1) = (4, -1)

Learn more about transformations here:

brainly.com/question/28354239

brainly.com/question/27743837

5 0

1 year ago

David is going to install tiles on the floor of two rooms. One of the floors is 18 feet long by 12 feet wide. The other floor is

crimeas [40]

Dimension of one of the floors of one room that David wants to install tiles is 18feet long by 12 feet wide
Then
Area of the above room = 18 * 12 square feet
   = 216 square feet
Dimension of the floor of the other room that David wants to install tiles is 24 feet long and 16 feet wide
Then
Area of the other room = 24 * 16 square feet
   = 384 square feet
Then
The total square feet of the
rooms that David wants to install tiles = 216 + 384
   = 600 square feet
Cost of the tile that covers 1 square feet = $5
Cost of the 4 tiles that cover 4 square feet = $17
Then
Area that can be covered with 4 square feet of tiles = 600/4 square feet
      = 150 square feet
Minimum cost of covering
the two rooms that David wants to install tiles = 150 * 17 dollars
      = 2550 dollars
So the minimum cost of installing the tiles on the two floors of David's two rooms is $2550. I hope the procedure is simple enough for you to understand.

8 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

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

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

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

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago