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Sliva [168]
3 years ago
13

The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds

. If a college football player is randomly selected, find the probability that he weighs between 170 and 220 pounds.
Mathematics
1 answer:
belka [17]3 years ago
8 0

Answer:

P(170

And we can find this probability with this difference:

P(-0.6

Step-by-step explanation:

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(200,50)  

Where \mu=200 and \sigma=50

We want to find the following probability:

P(170

And we can use the z score formula given by:

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

And using this formula we got:

P(170

And we can find this probability with this difference:

P(-0.6

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The bureau of labor statistics reported the consumer price index as 211.4 in december 2007, and 231.1 in december 2012. by what
Anvisha [2.4K]

The first step to determine the percentage of increase is to compute for the difference of consumer price index in value.

231.11 (consumer price index for 2012) – 211.4 (consumer price index for 2007) = 19.71 (increase)

To compute for the percentage of increase, you need to compare the increase to the consumer price index for 2007.

19.71/211.4 = 9.3% increase

5 0
3 years ago
What are the roots of the equation x^2+6x-9=10?
Ronch [10]

Answer:

-3+2sqrt7

-3-2sqrt7

Step-by-step explanation:

x^2+6x-9=10

x^2+6x-9-10=0

x^2+6x-19=0

ax^2+bx+c=0

a=1 b=6 c=-19

As cannot be solved by completing square we will use quadratic equation

x= (-b+sqrt(b^2-4ac))/2a     and      x= (-b-sqrt(b^2-4ac))/2a

x= (-6+sqrt(6^2-4*-19))/2     and      x= (-6-sqrt(6^2-4*-19))/2

x=(-6+sqrt(36+76))/2           and      x=(-6-sqrt(36+76))/2

x=(-6+4sqrt7)/2                   and       x=(-6-4sqrt7)/2

x=(-3+2sqrt7)                       and      x=(-3-2sqrt7)

x=2.29                                 and      x= -8.29

3 0
4 years ago
Trigonometry and trigonometric functions
zaharov [31]

Answer:

The trigonometric functions include the following 6 functions: sine, cosine, tangent, cotangent, secant, and cosecant. For each of these functions, there is an inverse trigonometric function. The trigonometric functions can be defined using the unit circle.

Step-by-step explanation:

6 0
3 years ago
The population of a town was 7652 in 2016. The population grows at a rate of 1.6% annually.
Serggg [28]

Answer:

(A) The population's growth rate in equation form is y = (0.016t * 7652) + 7652

(B)  y = (0.016t * 7652) + 7652 =

y = (0.016(8) * 7652) + 7652 =

y = (0.128 * 7652) + 7652 =

y = 979.456 + 7652 =

y = 8631.456 (Or About) 8631

Step-by-step explanation:

(A) Y = the total population of the town. 0.016 is 1.6% just in its original form. T = the year in which were trying to find the town's total population. 7652 is the total population of the town in 2016. With this information, the equation reads, The total population of the town (Y) is equal to 16% (0.016) of the current year's population (T) added to 2016's population of 7652. (This last sentence can also be read what is 1.6% of the towns population in the year were trying to find. Because the population is always growing, 1.6% gets multiplied as to scale with the total population in year T)

(B) We just substitute (T) for 2024, or 8 years after 2016 (2024-2016) and simplify the equation we made.

7 0
3 years ago
A frequency table for the 30 best lifetime baseball batting averages of all time is shown to the right. These data can be graphi
Mashcka [7]

Answer: B

Step-by-step explanation:

Histogram is a statistical graph with the use of bar. The bar are not seperated unlike bar chart.

0.320 to 0.329 and 0.360 to 0.369 are of the same frequency which is equal to one. 0.350 to 0.359 is of frequency 2 a little above frequency 1.

Option B and D are very close to each other in value representation. But the frequency of 0.350 to 0.359 in option D is 3. This renders option D invalid and make option B the correct answer.

5 0
3 years ago
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