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3 years ago

Given the values in the table on the left come from

Mathematics

1 answer:

schepotkina [342]3 years ago

7 0

–3.2 < x < –2.4

1.6 < x < 2.4

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PLSSSSSS HELP 100 POINTS What is this figure a picture of:

NARA [144]

Answer:

Step-by-step explanation:

Because i found the picture on goo gle and it said tessellation

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2 years ago

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Will bought a pizza from mazzio's. His 7 friends come over for a pizza party. Each friend gets 2 slices of pizza. after the pizz

Readme [11.4K]

Answer:

14/17

There were 17 before the party and 3 after ward.

Step-by-step explanation:

7 x 2 = 14 + 3 = 17

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3 years ago

1. Y = x + 6 Y = 2x A) (6, 12) B) (-12, -6) C) (-6,-12) D) (2,4)

LuckyWell [14K]

The answer is a, if x = 6 and y = 12, you get 6+6=12 and 2*6=12
Hope this helps!

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3 years ago

Whcih of these equations define y as a nonlinear function of x?

dexar [7]

Answer:

hello :

note :

the linear function of x have equation : y = ax + b

correct answer is : A) and C) and E)

A) : 3y = 2x -15

y = (2/3)x -15/3

y = ( 2/3)x -5 ..... when : a = 2/3 and b = -5

B) : 5(x+y)=-25

x + y = - 5

y = - x -5 ...... when : a = -1 and b = -5

C) y = 1

y = 0x +1 .....when : a = 0 and b = 1

8 0

4 years ago

The 2003 Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United Sta

levacccp [35]

Answer:

See explanation

Step-by-step explanation:

5 of 15 top-ranking restaurants offer dinner for more than $50 and 10 offer dinner for less than $50.

You will eat dinner at three of these restaurants.

a) The probability that none of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_3}{C^{15}_3}=\dfrac{\frac{10!}{3!(10-3)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 12!}{7!\cdot 15!}=\dfrac{8\cdot 9\cdot 10}{13\cdot 14\cdot 15}\approx 0.2637$

b) The probability that one of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_2\cdot C^5_1}{C^{15}_3}=\dfrac{\frac{10!}{2!(10-2)!}\cdot \frac{5!}{1!(5-1)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 5!\cdot 3!\cdot 12!}{2!\cdot 8!\cdot 4!\cdot 15!}=\dfrac{ 9\cdot 10\cdot 5\cdot 3}{\cdot 13\cdot 14\cdot 15}\approx 0.4945$

c) The probability that two of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_1\cdot C^5_2}{C^{15}_3}=\dfrac{\frac{10!}{1!(10-1)!}\cdot \frac{5!}{2!(5-2)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 5!\cdot 3!\cdot 12!}{9!\cdot 2!\cdot 3!\cdot 15!}=\dfrac{10\cdot 3\cdot 4\cdot 5}{13\cdot 14\cdot 15}\approx 0.2198$

d) The probability that all three of the meals will exceed the cost covered by your company is

$\dfrac{C^{5}_3}{C^{15}_3}=\dfrac{\frac{5!}{3!(5-3)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{5!\cdot 12!}{2!\cdot 15!}=\dfrac{3\cdot 4\cdot 5}{13\cdot 14\cdot 15}\approx 0.0220$

8 0

4 years ago