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nasty-shy [4]
3 years ago
15

Solve.

Mathematics
2 answers:
aivan3 [116]3 years ago
8 0
She spent 1.97 on each beach ball
pishuonlain [190]3 years ago
4 0

Number of beach balls Polly bought =12

Total cost of 12 beach balls = $23.64

Here in order to find cost of one beach ball, we will apply unitary method.

So to find cost of one beach ball we will divide cost of 12 beach balls by 12

So cost of one beach ball = 23.64 /12 = 1.97.

So cost of one beach ball is $1.97

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Answer:

1,100

Step-by-step explanation:

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A system of equations is graphed on the coordinate plane.
Nikolay [14]

Answer:

no solutions

Step-by-step explanation:

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PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
GrogVix [38]

Answer:

1. D

2. B

3. C

4. A

Step-by-step explanation:

just know when the sentence says "each" or "per" next to a number, there need to be an x next to it.

In other words, I kinda just winged it :P

4 0
2 years ago
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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
3 years ago
-(x-5)^(1/4)+(7/3)=2 solve for x
Arisa [49]
I think the answer is 406/81. 

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4 years ago
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