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Rama09 [41]
2 years ago
12

White shapes and black shapes are used in a game some of these shapes are circles and all the rest of the shapes are squares the

ratio of the the white shapes to black shapes is 5:11 the ratio of white circles to white squares is 3:7 the number of black circles to black squares is 3:8 work out what shape of the shapes are circles?
Mathematics
1 answer:
daser333 [38]2 years ago
7 0

Answer:

(D) THE AREA CAN BE FOUND BY COURITING THE GRID SQUARES INSIDE THE SHAPE.

Step-by-step explanation:

:D

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Y=2/3x+2/3 bc slope is 2/3 and 4-2/4-1=2/3
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3 years ago
PLEASE HELP WILL GIVE BRAINLIEST.
kari74 [83]

Answer:

Choice D

Step-by-step explanation:

For this one I would find if the point lands on the line.

<em><u>Choice A:</u></em>

What we have to do is to plug in -4 for x and 4 for y.

y=2x+10\\(4)=2(-4)+10\\4=-8+10\\4\ne2

The point is not on this line so this cannot be it.

<em><u>Choice B:</u></em>

We pug what we know again.

y=\frac{1}{3}x+1\\(4)=\frac{1}{3}(-4)+1\\(4)=-\frac{4}{3}+1\\4\ne-\frac{1}{3}

The point is not on this line so it can't be it.

<em><u>Choice C:</u></em>

We pug in what we know again.

y=3x-9\\(4)=3(-4)-9\\(4)=-12-9\\4\ne-21

The point is not on this line so it can't be it.

The next one has to be it, but we'll check it just in case.

<em><u>Choice D:</u></em>

We plug in what we know again.

y= \frac{1}{2}x+6\\(4)=\frac{1}{2}(-4)+6\\(4)=-2+6\\4=4

The point is on this line so this is the line.

3 0
3 years ago
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb
ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

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25 ! i hope this helps :)
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