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sergiy2304 [10]
3 years ago
10

One Step Equations (Multiplication) -4=2y

Mathematics
1 answer:
Katena32 [7]3 years ago
8 0
The answer will be y= -2.
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Given that pentagon, ABCDE pentagon FGHIJ find the measure of <1
Art [367]

Answer:

d

Step-by-step explanation:


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A recent poll of 80 randomly selected Californians showed that 38% ( = 0.38) believe they are doing all that they can to conserv
Elanso [62]
The margin of error given the proportion can be found using the formula

z* \sqrt{ \frac{p(1-p)}{n} }

Where
z* is the z-score of the confidence level
p is the sample proportion
n is the sample size

We have
z*=2.58
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n=80

Plugging these values into the formula, we have:
2.58 \sqrt{ \frac{0.38(1-0.38)}{80} } =0.14

The result 0.14 as percentage is 14%

Margin error is 38% ⁺/₋ 14%
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3 years ago
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PLS HELP! Tanisha is planning a party. She will serve hamburgers, potato salad, strawberry shortcake, and lemonade. Including Ta
Fittoniya [83]
Shee needs 448 ounces of lemonade.

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80 ounces of lemonade needed 8 lemons.
448 ounces of lemonade needed 8 x 5.6 = 44.8 lemons.

Tanisha has 38 lemons. 
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4 0
3 years ago
James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo
Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒  for n = 2 , y = 4       ..... (2)

⇒y  = 2^2  =  2^n

Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

⇒  There are total 128 equal parts.

Lastly,  if the paper is folded ELEVEN  times  ⇒  n = 11

So, y = 2^ n = 2^{11} = 2048

⇒  There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

8 0
4 years ago
What is the radius and diameter of the following circle
brilliants [131]

Answer:

Radius: 16cm

Diameter: 32cm

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3 years ago
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