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Murljashka [212]
3 years ago
15

16×4=(7×4)+(10×4)=please help

Mathematics
2 answers:
mote1985 [20]3 years ago
5 0
False

16×4=64

7×4=28 and 10×4=40. (28+40)=68

68≠64
stealth61 [152]3 years ago
4 0
<span><span><span><span><span><span>(16)</span><span>(4)</span></span><span>=</span></span><span>(7)</span></span><span>(4)</span></span>+<span><span>(10)</span><span>(4)</span></span></span><span><span>64≠</span>68</span>False<span>MAYBE THIS?</span>
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The surface area of a cube is 600 cm. what is its volume? 200 1000 600 800​
Studentka2010 [4]

Answer:

1000 cm^3

Step-by-step explanation:

First, Find the side length of the cube. The surface area is all the area on the surface of a 3-dimensional object. Since a square has 6 congruent faces the formula for surface area is 6s^2 where s is the side length. Since we know the surface area we can solve for the side length.

600=6s^2- DIvide by 6

100=s^2- Take the square root of both sides.

10=s

The volume of a cube is s^3.

10^3=V

1000=V

3 0
2 years ago
A caterpillar moves at a constant speed of 1/2 inches per second. Let x represent the time travel in seconds and y represent the
Anika [276]

Step-by-step explanation:

Expression = 1/2x = y

Easy algebraic expression

~Done~

5 0
2 years ago
Three times each day, a quality engineer samples a component from a recently manufactured batch and tests it. Each part is class
Tresset [83]

Answer:

Step-by-step explanation:

Hello!

Three samples of components manufactured are taken per day. They are classified as:

D: "Conforming (suitable for its use)"

E: "Downgraded (unsuitable for the intended purpose but usable for another purpose)"

F: "Scrap (not usable)"

This classification includes the three events that may occur in your sample space S. The experiment consists in recording the categories of the three parts tested in a day.

a. List the 27 outcomes in the sample space.

The possible outcomes in the space sample are the combinations of the three events. To avoid using the same letters as in the following questions I've named the evets as D, E, and F

S={DDD, DED, DFD, DEF, DFE, DEE, DFF, DDE, DDF , EDE, EEE, EFE, EED, EEF, EDF, EFD, EDD, EFF , FDF, FEF, FFF, FFE, FFD, FDE, FED, FDD, FEE}

b. Let A be the event that all the parts fall into the same category. List the outcomes in A.

A: "All the parts fall into the same category"

You have three possible outcomes for this event, that the three compounds are conforming, "DDD", that the three are unconforming, "EEE", or that the three compounds are scrap, "FFF". There are only three possible outcomes for this event.

S={DDD, EEE, FFF}

c. Let B be the event that there is one part in each category. List the outcomes in B.

B: "There is a part in each category"

This means, for example, The first one is conforming "D", the second one is unconforming "E" and the third one is scrap "F", then the first one may be unconforming "E", the second one is conforming "D" and the thirds one is scrap "F", and so on, you have 6 possible outcomes for this event:

S={DEF, DFE, EDF, EFD, FDE, FED}

d. Let C be the event that at least two parts are conforming. List the outcomes in C.

C: "At least two parts are conforming"

For this event, you can have two of the compounds to be considered conforming or the three of them.

S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

A total of 7 combinations fit this event.

I hope you have a SUPER day!

6 0
3 years ago
Points A, B, C, and D form the rectangle ABCD. E, F, and H are three points of another rectangle EFGH. Where does the fourth poi
ycow [4]

Answer:

D. (7.4); scale factor 2.

I’m not sure how to explain this but I’ve done it last year.

Hope that helps.

5 0
3 years ago
180kg, 250kg, 200kg, 209kg, 195kg, 205kg, 190kg, 188kg, 192kg. What is the IQR of the lion's weights.
velikii [3]

Answer:

18

Step-by-step explanation:

Given the data:

180kg, 250kg, 200kg, 209kg, 195kg, 205kg, 190kg, 188kg, 192kg

The interquartile range (IQR) = Q3 - Q1

Reordering the data:

180, 188, 190, 192, 195, 200, 205, 209, 250

Q3 = 3/4(n+1)th term

Q3 = 3/4(10) = 7.5 th term = (205+209)/2 = 207

Q1 = 1/4(n+1)th term

Q1 = 1/4(10) = 2.5th term = (188+190)/2 = 189

Q3 - Q1 = 207 - 189 = 18

3 0
3 years ago
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