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Alchen [17]
3 years ago
8

what is 48 35 26 31 27 52 55 63 18 74 35 61 28 42 76 35 38 64 71 26 53 65 47 41 102 38 35 84 41 26 73 62 35 39 62 49 104 62 54 1

9 28 64 36 39 53 89 26 53 43 75 35 39 29 26 45 39 31 30 70 42 37 54 85 42 36 91 25 28 53 65 78 42 68 44 41 50 46 48 75 29 53 72 17 28 38 30 48 43 27 46 36 48 62 31 42 48 61 22 124 27 53 86 60 47 41 37 27 36 45 80 45 70 53 25 29 16 30 36 38 42 48 51 38 39 43 61 75 38 42 55 72 39 28 51 43 49 67 25 37 39 46 41 50 93 53 82 47 36 38 51 42 40 24 68 38 36 34 49 65 68 62 53 67 23 51 40 43 49 57 61 26 54 35 32 38 39 24 added all together
Mathematics
1 answer:
Montano1993 [528]3 years ago
6 0

Answer: 7,824

Step-by-step explanation:

What kind of question is this lol, it took me forever.

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Can someone please help me
Mashutka [201]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Functions.

Let's first find the g(-1),

so we get as

3(-1)^2 +5(-1)-6

=> 3 -5-6

=> -8

now since g(-1) =-8

let's find f(g(-1)) that is f(-8)

f(-8) = 4(-8) + 14

=> f(g(-1)) = -32+14

=> f(g(-1)) = -18

-18 is the answer.

8 0
3 years ago
What is the distance between (1, 3) and (5, 8)
katovenus [111]
I think you’d do (5,8) - (1,3) to get (4,5)
8 0
3 years ago
Read 2 more answers
A sports broadcaster predicts a basketball team will score 88 point me in the next game. If the team actually scored 98 points w
gavmur [86]

Answer:

The percentage error in the prediction = 10.20%

Step-by-step explanation:

A sports broadcaster predicts a basketball team will score 88 point me in the next game.

If the team has actually scored 98 points we have to find the percentage error in the broadcaster's prediction.

The deviation of the prediction from the actual value = 98 - 88 = 10.

The formula for finding error percentage = \frac{deviation from actual value}{prediction}

                                       = \frac{10}{98} \times 100

                                       = 10.20 %

The percentage error = 10.20%

5 0
3 years ago
6.1.3
Nata [24]

Answer:

The requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

Step-by-step explanation:

A normal-distribution is an accurate symmetric-distribution of experimental data-values.  

If we create a histogram on data-values that are normally distributed, the figure of columns form a symmetrical bell shape.  

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variates is known as the standard normal distribution.

Thus, the requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

8 0
3 years ago
Can y’all please help me ?
hjlf

Answer:

4.3

Step-by-step explanation:

6 0
3 years ago
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