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2 years ago

Can u pls help me with this

Mathematics

2 answers:

sp2606 [1]2 years ago

8 0

I believe there is one solution as 10/7 and -7/4 can be solved and we can get the value of y.

jarptica [38.1K]2 years ago

5 0

I believe it doesn’t have a solution because even if the 9/7x are the same, the other fractions are not the same value therefore, there is no solution since
+10/7≠ -7/4

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Given R (7,-1), A (3,-6), B(-3,-6), E (-5,4), plot the points and trace the figure.

RUDIKE [14]

Answer:

See below

Step-by-step explanation:

Given R (7,-1), A (3,-6), B(-3,-6), E (-5,4)

Points plotted and they form a quadrilateral

See attached

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3 years ago

3/4 of a class went to see a movie. If 1/2 of the class went to the afternoon session, what fraction of the class went to the ev

Paha777 [63]

1/4 went to the evening session

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3 years ago

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5) The graph of the parent linear function f(x) = x is shown in black on the coordinate grid. Write the function that represents

djyliett [7]

Answer:

(m) increased, (b) unchanged. g(x)

(m) decreased, (b) unchanged. m(x)

(m) unchanged, (b) increased. h(x)

(m) unchanged, (b) decreased. n(x)

f(x): m = -1/2 b = 2

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2 years ago

Find the percent of increase or decrease?<br> N to 4/5n?

d1i1m1o1n [39]

Answer:

Decrease by 20%.

Step-by-step explanation:

We have to find the percent of increase or decrease from n to $\frac{4}{5}n$ .

Now, $\frac{4}{5}n = 0.8n < n$

So there is a decrease fro n to $\frac{4}{5}n$ .

Now, the percentage of decrease will be given by

= $[ \frac{\textrm {Original Value - Final Value}}{\textrm {Original Value}}]\times 100 \%$

= $[ \frac{n - 0.8n}{n}]\times 100 \%$

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3 years ago

Identify the class width, class midpoints, and class boundaries for the given frequency distribution.

Crank

Answer:

a. Class width=4

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

$M=\frac{lower class limit+upper class limit}{2}$

Class

Interval Midpoints

45-48 $\frac{45+48}{2}=46.5$

49-52 $\frac{49+52}{2}=50.5$

53-56 $\frac{53+56}{2}=54.5$

57-60 $\frac{57+60}{2}=58.5$

61-64 $\frac{61+64}{2}=62.5$

65-68 $\frac{65+68}{2}=66.5$

69-72 $\frac{69+72}{2}=70.5$

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0

3 years ago