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Setler79 [48]
3 years ago
12

Solve 1 5/7 • x for x = 5/6

Mathematics
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

7/18

Step-by-step explanation:

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I am confused help??
PtichkaEL [24]

Answer:

48

Step-by-step explanation:

2 x 2 x 2 x 2 = 16

16 + ?

4 x 4 x 4 = 64

0.5 x 64 = 32

32 + 16 = 48

5 0
2 years ago
Read 2 more answers
How many different ways can you find for 7 coins totaling 95 cents?​
MatroZZZ [7]

Answer:

one way

Step-by-step explanation:

2 quarters

4 dimes

1 nickel

3 0
3 years ago
The average annual income, I, in dollars, of a lawyer with an age of x years is modeled with the following function:
Eduardwww [97]
I = -425x^2 + 45500x - 650000
For I to be maximum, dI/dx = 0
dI/dx = -850x + 45500 = 0
x = 45500/850 = 53.53

Therefore, maximum I = -425(53.53)^2 + 45500(53.53) - 650000 = 567,794.11

Therefore, <span>the maximum average annual income, in dollars, a lawyer can earn</span>  is $567,794
8 0
4 years ago
Where can I contact the CEO of brainly I feel like they should allow more then 2 answers per question to create a more variety o
Alexandra [31]

Answer:

thats why they put premium

Step-by-step explanation:

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8 0
3 years ago
A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a
natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

The sample means  = x₁  +  x₂  +  .   .   .  , x₂₃ / 23

= \frac{x_1 +x_2 +...x_2_3}{23}

= 2.3hr

⇒\sum xi =2.3 \times 23 = 52.9hrs

let x₁ , x₂, .   .   .  , x₂₃  arranged in ascending order

Then x₂₃ was 10  and has been changed to 14

i.e x₂₃ increase to 4

Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

therefore, the new sample mean is 2.47

2) For the old data set

the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = [{x_1_2]^t^h value remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0
4 years ago
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