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jeka57 [31]
3 years ago
14

1 For each line

Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

Step-by-step explanation:

I'll do line A for you and you can use the formulas to solve lines B and C yourself, since its good for you to practice doing these questions yourself

a)

The gradient, m, is calculated using m = (y2-y1)/(x2-x1) where x1,x2,y1 and y2 can be any ordered pairs on the line.  I'm going to use (4,0) and (7,3) as the 2 points.

m = (3-0)/(7-4) = 3/3 = 1

b)

The y-intercept is where the line intersects with the x-axis. In this case (0,-4)

c)

The equation of a linear line is y=mx+b (or c depending on which country you are from)

y = 1x-4

y=x-4

Now try the other 2 lines yourself!

If this answer has helped you, considered making this the brainliest answer!

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there are 48 students in a school play. the ratio of boys to girls is 5:7. how many more girls than boys are in the school play?
Alex Ar [27]
We can use trial or error method here...

If we take 4 table, 4*5 = 20 and 4*7 = 28. 20+28 = 48.
20 : 28 = 5 : 7

So, there are 20 boys and 28 girls in a school play and there are 8 girls more than boys.

If you look over tables and try it out, only 4 table works for the given ratio.
3 0
3 years ago
1 MINUTE!!!
iris [78.8K]

Answer:

About 9.38

Step-by-step explanation:

Since the \sqrt{88} is between the \sqrt{81} and the \sqrt{100} which simplified is 9 and 10, you just need to guess and check. Eventually you'll get about 9.38 (I used Google cause I was lazy) And if it isn't between the numbers 9 & 10, it's wrong.

6 0
3 years ago
Read 2 more answers
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
Can someone please answer these ?!
katrin [286]

Answer: b and d

Step-by-step explanation:

7 0
3 years ago
Solve the inequality and enter your solution as an inequality. <br><br> 8 + x&gt;4​
castortr0y [4]

if any concerns or questions on solution, just comment

6 0
2 years ago
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