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diamong [38]
2 years ago
12

Find m Given The problem

Mathematics
1 answer:
solmaris [256]2 years ago
5 0

Answer:

75

Step-by-step explanation:

Its an acute

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Tell me how to shade the area on the grid that shows 7/8x3/4
viva [34]
If the grid is an 8x8: 

To solve this we need to convert 3/4 into a fraction x/8 so that they have the same denominator.

To do this we can set a porportion and find a common denominator. 4 x 2 = 8 so the common denominator would be 8 and the rule would be x 2 so 4 x 2 = 8 and 3 x 2 = 6

So we get 7/8 and 6/8

From top to bottom shade 7 spaces in a row

From left to right shade 6 spaces in a row

Fill in the spaces in between to make a square

I hope this helps!
8 0
3 years ago
Read 2 more answers
A certain plant grows 1 3/4 inches every week how long will it take the plant to grow 9 1/4 inches. HELPPP I’m dumb
GREYUIT [131]
It only takes 13/4 days inn week
3 0
3 years ago
Please help asap!! Due today!
Elena-2011 [213]

Answer:

25 cm

Step-by-step explanation:

a^2 + b^2 = c^2

7^2 + 24^2 = c^2

49 + 576 = c^2

c^2 = 625

c = 25

Answer: 25 cm

5 0
2 years ago
On New years Eve, the temperature fell by 24 degrees in a span of 6 hours. By how many degrees did the temperature fall every ho
DochEvi [55]

Answer:

4

Step-by-step explanation:

24/6

3 0
3 years ago
Suppose a normal distribution has a mean of 222 and a standard deviation of 16. What is the probability that a data value is bet
valentina_108 [34]
Mean of the distribution = u = 222
Standard Deviation = s = 16

We have to find the probability that a value lies between 190 and 230.

First we need to convert these data values to z score.

z= \frac{x-u}{s}

For x = 190, 
z= \frac{190-222}{16}=-2

For x = 230
z= \frac{230-222}{16}=0.5

So, we have to find the percentage of values lying between z score of -2 and 0.5

P( -2 < z < 0.5) = P(0.5) - P(-2)

From standard z table, we can find and use these values.

P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687

Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution. 

5 0
3 years ago
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