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slavikrds [6]
3 years ago
15

Can someone help me? Please and thank you !

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0
FOR question 11 use the inverse to find the joint relative frequency. So do 98-60.
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The volume of a rectangular prism is 2,058 cubic cm. The length of the prism is 3 times the width. The height is twice the width
liraira [26]
Let the width be w, length = 3w and height = 2w

Volume = length x width x height = w x 3w x 2w = 6w^3
6w^3 = 2,058
w^3 = 2,058/6 = 343
w = ∛343 = 7

width = 7 cm
8 0
4 years ago
The mean amount spent by a family of four on food is $500 per month with a standard deviation of $75. Assuming that the food cos
Alexeev081 [22]

Answer:

z=\frac{x- \mu}{\sigma}

And replacing we got:

z=\frac{425-500}{75}= -1

And we can calculate this probabilit using the normal standard distribution or excel and we got:

P(z

Step-by-step explanation:

If we define the random variable of interest "the amount spent by a family of four of food per month" and we know the following parameter:

\mu = 500, \sigma = 75

And we want to find the following probability:

P(X

And we can use the z score formula given by:

z=\frac{x- \mu}{\sigma}

And replacing we got:

z=\frac{425-500}{75}= -1

And we can calculate this probabilit using the normal standard distribution or excel and we got:

P(z

5 0
3 years ago
an art student wants to take a triangle with side lengths of 9,9, and 15 and enlarge it so that the longest side of a new triang
VikaD [51]

Answer:

13.5

Step-by-step explanation:

The longest side of the smaller triangle is 15 while that of the bigger one is 22.5

The enlargement ratio is 22.5/15 = 1.5

so the lengths of the smaller sides will be;

(1.5*9) = 13.5

6 0
3 years ago
Why are circles so important in creating equilateral triangles
Daniel [21]
Maybe because the diameter of the circle should equal each side length of te equilateral triangle 
6 0
3 years ago
Okay so I NEED help tonight. Please someone I really have to finish this tonight.
nlexa [21]
What do you need to finish
6 0
3 years ago
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