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sp2606 [1]
3 years ago
5

12 of 18

Mathematics
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

3x+3y and distributive property

Step-by-step explanation:

It is given that

Gretchen buys three bags of apples = 3X

Hayley buys three bunches of bananas = 3Y

As it is also mentioned that it first it is multiplied and then it is added

So, the expression for determining the total number of fruit pieces is

3X + 3Y

And, this expression denotes the distributive property in which we multiplied the each variable and then added it

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PLS HELP ME ASAP WITH THIS WHOLE PAGE! THANK YOU SOO MUCH!!
Doss [256]
1a) The function has arrows on both ends and no place in the middle where it is not defined. Its domain is ...
  All Reals

1b) The function gives no output values below -3, but it gives output values of -3 and all above that. Its range is ...
  y ≥ -3

1c) For values of x less than -1, the function's output is 1. This matches g(x) and s(x). At x=0, the function's output is -3, which only matches g(x). The appropriate choice is ...
  g(x)


2b) The function is only defined for 0 ≤ x < 8. This is its domain.

3) A definition might be ...
y=\left\{ \begin{array}{rcl}x+1 & \mbox{for} & -4\le x \le 1\\-x & \mbox{for} & 1 < x \le 3 \end{array}

6 0
3 years ago
Form a sequence that has two arithmetic means between -13 and 89. a. -13, 33, 43, 89 c. -13, 21, 55, 89 b. -18, -36, -72, -144 d
iogann1982 [59]
<span>Form a sequence that has two arithmetic means between -13 and 89. a. -13, 33, 43, 89 c. -13, 21, 55, 89 b. -18, -36, -72, -144 d. -18, -81, -144

Solution:

Since  it has to be between -13 and 89, letter d and b are not anymore considered to be the answer.

for a:
33-(-13)=46=d
43-33=10=d the value for this d is different from the two sequence,
89-43=46=d
they have different value for d, thus this is not the answer!

for c:
21-(-13)=34=d
55-21=34=d
89-55=34=d

they have the same value for d, thus the correct answer is </span><span> c. -13, 21, 55, 89</span>
8 0
3 years ago
15600=x (x+1)(x+2) What is x?
Andrew [12]
The answer is x = 24
4 0
3 years ago
Read 2 more answers
PLS HELP SHOW ALL YOUR WORKING OUT BRAINLIEST! WILL BE GIVEN :D
aliina [53]

Answer:

(a) Five

(b) Thirteenth

Step-by-step explanation:

We have the mean is 12.6 and

that 11 occurs 4 times, 12 occurs 7 times, 13 occurs 9 times, and 14 occurs f times.

How many children are there in all (that is the sum of the frequencies).

4+7+9+f

20+f

Alright so the mean could be found by doing:

\frac{4(11)+7(12)+9(13)+f(14)}{20+f}

But we are given this is also equal to: 12.6.

So we have

\frac{4(11)+7(12)+9(13)+f(14)}{20+f}=12.6

I'm going to simplify what I can on top.

\frac{245+14f}{20+f}=12.6

I'm going to write 12.6 as \frac{12.6}{1} because I want to cross multiply:

\frac{245+14f}{20+f}=\frac{12.6}{1}

(245+14f)(1)=12.6(20+f)

Distribute:

245+14f=252+12.6f

Subtract 12.6f on both sides:

245+1.4f=252

Subtract 245 on both sides:

1.4f=7

Divide both sides by 1.4:

f=5

There are five 14 years old.

(b) I would say 13 is good age to represent this bunch.

The means was 12.6 which when rounded is 13.

The mode is 13 because it is the most occurring

The median is also 13. Why? If you list out the data 13 will be the middle number. Or you could say there are 25 kids and if I divide it by 2, I get 12.5.  This means you only need to count to the 13th kid with the ages in order to tell with the median is.

There are 4 eleven yr olds.

There are 7 twelve yr olds.  That is 11 kids so far.

So the median has to be included in the 9 thirteenth yr olds.

3 0
3 years ago
A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin
Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

P(\dfrac{X}{Y}) = 60\% = 0.6

P(\dfrac{X}{Y'}) = 0.01\% = 0.0001

P(Y) = 0.01

Thus, P(Y') \ will\ be = 1 - P(Y)

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

P(YX) = P(\dfrac{X}{Y}) \ P(Y)

P(YX) =0.6 \times 0.01

P(YX) =0.006

The probabilities of not involved in cheating & the evidence are present is:

P(Y'X) = P(Y')  \times P(\dfrac{X}{Y'})

P(Y'X) = 0.99  \times 0.0001 \\ \\  P(Y'X) = 0.000099

(b)

The required probability that the evidence is present is:

P(YX  or Y'X) = 0.006 + 0.000099

P(YX  or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}

P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}

P(\dfrac{Y}{X}) = 0.9838

5 0
3 years ago
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