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Mama L [17]
2 years ago
10

A store manager is looking at past jewelry sales to determine what sizes of rings to keep in stock. The list shows the ring size

s purchased by the last ten jewelry customers.
9, 7, 6.5, 7.5, 7, 8, 5, 6, 7.5, 8
What is the variance of the data? Round to the nearest hundredths.
Mathematics
2 answers:
11Alexandr11 [23.1K]2 years ago
5 0

Answer: 1.15 is the correct answer or C on edg.

Step-by-step explanation:

kirza4 [7]2 years ago
4 0
Variance is the average of the square of the differences of each data with the mean. To calculate for the variance, we first calculate for the mean. Then, we subtract each data with the mean. Next, each difference would be squared and added. The resulting value would be divided on how many data are used. We calculate as follows:

Mean = <span>9 + 7 + 6.5 </span>+ 7.5 + 7 + 8 + 5 + 6 + 7.5 + 8 / 10
Mean = 6.4

Squared of the sum of the differences = (9-6.4)^2 + (7-6.4)^2 + (6.5-6.4)^2 + (7.5-6.4)^2 + (7-6.4)^2 + (8-6.4)^2 + (5-6.4)^2 + (6-6.4)^2 + (7.5-6.4)^2 + (8-6.4)^2 = 17.15

Variance = 17.15 / 10 = 1.715
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30 POINTS! Monica paid sales tax of $1.50 when she bought a new bike helmet. if the sales tax rate was 5%, how much did the stor
avanturin [10]
Hmm... well, here ya go. 

1.50 ÷ 0.05 = 30 
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Hope I helped.  
6 0
2 years ago
Read 2 more answers
The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced
Dahasolnce [82]

Answer:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.86,0.13)  

Where \mu=5.86 and \sigma=0.13

We are interested on this probability

P(5.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

4 0
2 years ago
The acceleration of an object due to gravity is 32 feet per second squared. What is acceleration due to gravity in inches per se
GarryVolchara [31]
For each foot there are 12 inches  so we multiply by 12

32*12 = 384 inches

C
5 0
2 years ago
Read 2 more answers
The quality control team of a company checked 800 digital cameras for defects. The team found that 20 cameras had lens defects,
dmitriy555 [2]

Answer:

0.025

Step-by-step explanation:

-This is a conditional probability problem.

-Let L denote lens defect and C charging defect.

#We first calculate the probability of a camera having a lens defect;

P(lens)=\frac{Lens}{Total}\\\\=\frac{20}{800}\\\\=0.025

#Calculate the probability of a camera having a charging defect:

P(Charging)=\frac{Charging}{Total}\\\\=\frac{25}{800}\\\\=0.03125

The  the probability that a camera has a lens defect given that it has a charging defect is calculated as:

P(L|C)=\frac{P(C)P(L)}{P(C)}\\\\=\frac{0.025\times 0.03125}{0.03125}\\\\=0.025

Hence,  the probability that a camera has a lens defect given that it has a charging defect is 0.025

6 0
2 years ago
Three percent of Jennie’s skin cells were burned when she escaped from a fire. If 3.7×10’(of her skin cells were burned then, ho
lora16 [44]

Answer:

The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  .

Step-by-step explanation:

Given as :

The percentage of Jennie's skin cell were burned = 3 %

The quantity of Jennie's skin cell were burned = 3.7 × 10^{10}

Let The quantity of Jennie's skin cell were not burned = x

Let Total quantity of whole skin cell = n

<u>Now, According to question</u>

The quantity of Jennie's skin cell were burned = Total quantity of whole skin cell × percentage of Jennie's skin cell were burned

i.e 3.7 × 10^{10} = n × 3 %

Or, 3.7 × 10^{10} = n × \dfrac{3}{100}

Or, n =  3.7 × 10^{10} × \dfrac{100}{3}

Or, n = \frac{3.7\times 10^{12}}{3}

i.e n = 1.233 × 10^{12}

Or, Total quantity of whole skin cell = n = 1.233 × 10^{12}

<u>Now, Again</u>

∵ percentage of Jennie's skin cell were burned = 3 %

So, percentage of Jennie's skin cell were not burned = 100 % - 3 % = 97 %

so , The quantity of Jennie's skin cell were not burned = 97 % of Total quantity of whole skin cell

Or, x = \frac{97}{100} × n

Or, x =  \frac{97}{100} × 1.233 × 10^{12}

∴ x = 119.601 × 10^{10}

So, The quantity of Jennie's skin cell were not burned = x = 119.601 × 10^{10}

Hence,  The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  . Answer

6 0
2 years ago
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