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Westkost [7]
3 years ago
11

a science center has v visitors on the first day of an special exhibit and 200 visitors on the second day what is the average nu

mber of visitors per day over the two days​
Mathematics
1 answer:
Luda [366]3 years ago
6 0
(V + 200) divided by 2= average of the 2 days
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How do you Order the numbers from least to greatest. -7 , -7.75, 1/2, 75%
Serggg [28]

Answer:

-7.75, -7, 1/2, 75%

Hope this helps ☝️☝☝

6 0
3 years ago
A car traveling at a certain speed will travel 76 feet per second. How many miles will the car travel in 3.1 hours if it maintai
murzikaleks [220]
\sf  \frac{76~ feet}{1~second} \times  \frac{1~mile}{5280~feet} \times  \frac{3600~seconds}{1~hour} =  \frac{51.82~miles}{hour}

51.82 miles in one hour, so 51.28 x 3 = <span>155.46 miles in 3 hours.

Your answer is </span><span>155.46 miles in 3 hours.</span>
6 0
3 years ago
If f(x) = 2x - 4 and g(x)= x², what is f(g(x))?
kupik [55]

Answer:

D

Step-by-step explanation:

To evaluate f(g(x)) substitute x = g(x) into f(x)

f(g(x))

= f(x²)

= 2(x²) - 4 = 2x² - 4

6 0
3 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
If n is a positive integer and the product all the integeres from 1 to n inclusive is a multiple of 990 what is the least possib
kolbaska11 [484]
Hello,

Since 990=2*3²*5*11 ==>n=11
6 0
3 years ago
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