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Ivenika [448]
3 years ago
13

Two spinners have been spun 100 times. Using the experimental probabilities below, what is the best guess for the number of time

s these spinners would land on the same number in 10 spins?
Mathematics
1 answer:
Agata [3.3K]3 years ago
3 0
It could be any time, but I would say 10. 10*10=100

hope it helps!
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On an algebra test , the highest grade was 50 points higher than the lowest the sum of the two grades was 122 find the lowest gr
KATRIN_1 [288]

Answer:

Highest grade: 86

Lowest grade: 36

Step-by-step explanation:

Let the highest grade be h, lowest grade be l.

h = 50 + l  \\  \\ h + l = 122 \\ h = 122 - l \\  \\

Therefore,

122 - l \:  = 50 + l \\ 2l \:  = 72 \\ l = 36 \\  \\ h = 36 + 50 = 86

6 0
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pentagon [3]

Answer:

D. f(x)=4x^{2}+1

Step-by-step explanation:

There is a translation 1 point up along the y axis and a compression of 4.

Moving a function up (let's use <em>h</em> for the amount of points up) would change the function as so:

f(x)=x^{2} +h

Meanwhile, the compression would modify x in this case. You can eliminate any answers (A. and B.) that have no modification to x, and eliminate C., as a fraction modification would actually widen the graph instead of compress it.

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4 0
3 years ago
How do you do this question?
daser333 [38]

Answer:

B. 1/2

Step-by-step explanation:

\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}

If we plug in 0 for z, we get 0/0.  Apply l'Hopital's rule.

\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}

Now when we plug in 0 for z, we get:

\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}

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kherson [118]
Here is your answer. Good luck!

6 0
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Ierofanga [76]

Answer:

Natalie saved 16%

Step-by-step explanation:

Natalie saved 16% because 20 divided by 125 is 0.16

When we find percentages our goal is to get something like 0.24 which would mean 24% or 0.14 or 14% in this case it is perfect because we immediately see 0.16 or 16% leaving us with our answer, Natalie got 16% off her new phone.

6 0
3 years ago
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