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iVinArrow [24]
2 years ago
15

Answer the question please

Mathematics
2 answers:
Soloha48 [4]2 years ago
7 0
Its jane because every hour, she earns $12
oksian1 [2.3K]2 years ago
5 0
Jane because she earns £12 for every hour that she works.
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Find the common ratio r for the geometric sequence and use r to find the next three terms. Enter your answer to four decimal pla
kirza4 [7]

Answer:

the common ration is -0.4. Three terms: 2.56,-1.024, and 0.4096
Step-by-step explanation:

You just keep on multypling the terms by -0.5 ,and you get.

7 0
2 years ago
Can someone help me on this math problem ?
faltersainse [42]

Answer:

4/3

Step-by-step explanation:

Given two points, we can find the slope by using

m = (y2-y1)/(x2-x1)

    = (20 - -16)/(7 - -20)

    = (20+16)/(7+20)

    = 36/27

   =4/3

8 0
3 years ago
Can You Solve This?<br> 6 = 2(1+2) =
nata0808 [166]

Answer: 6 = 2(1+2), this is true.

Step-by-step explanation:

To solve this, 6 = 2(1 +2) =

This question is to proove that 2(1 +2) = 6

First we open the bracket:

(2 × 1) + (2 × 2)

Multiply 2 by 1 = 2

Multiply 2 by 2 = 4

2 + 4 = 6

I hope this helps.

8 0
3 years ago
Read 2 more answers
If A parking garage charges three dollars for every hour how much would the cost be to park for t hours
SSSSS [86.1K]

Answer:

3h

hope this helps

have a good day :)

Step-by-step explanation:

3 0
2 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
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