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Orlov [11]
3 years ago
7

Enter the rule (equation) for the table: Answer: y=

Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0
Y is equals to 3

Explanation:
1+3 is 4 and 4+3 is 7 and it continues
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Alenkinab [10]
I’m pretty sure the anser is b
3 0
3 years ago
Read 2 more answers
Suma de fracciones <br> 5/8 + 3/4=
Finger [1]

Answer:

11/8

Step-by-step explanation:

5/8 + 3/4

        x2

5/8 + 6/8 = 11/8

5 0
3 years ago
Simplify 7p³ + 2m³ - 6m³ + 4p - 9p + 6p³
krok68 [10]
7p³ + 2m³ - 6m³ + 4p - 9p + 6p³
= 13p³ - 4m³ - 5p
= -4m³ + 13p³ - 5p

answer
d) -4m³ + 13p³ - 5p
6 0
3 years ago
Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou
monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )

                   = P ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) = P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )

                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = P (\frac{6.2-8}{1.5} <  \frac{x - u}{ std} < \frac{7-8}{1.5} )

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0
4 years ago
Tell me your cringes pick up lines
Bond [772]

Answer:

Are you a speeding ticket? Because you look fine.

Yikes, that was cringey

4 0
2 years ago
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