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Effectus [21]
4 years ago
5

seth has 23 u.s stamps and 14 foreign stamps he has 27 u.s coins ans 11 foreign coins what fraction of seths collection is from

the u.s write the fraction in simplest form
Mathematics
1 answer:
Hatshy [7]4 years ago
3 0
23+27=50 US collection
14+11= 25 Foreign collection
There is a total of 75 (25+50=75) in the entire collection.
And 50 of them are from the US

50/75=2/3
2/3 is the simplest form.

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On a workday the average decibel level of a busy street is 69 db, with 139 cars passing a given point every minute. if the numbe
Vlad1618 [11]
Decibel is directly proportional to number of cars passing the point. That is,

69 db ----- 139 cars
x   db ----- 17 cars

x = number of cars passing the point per minute on a weekend.

Then,
x= (69*17)/139 = 8.43 db ≈ 8 db
8 0
3 years ago
Please help its very urgent toooooooo .
GenaCL600 [577]

Answer

240 degrees

Step-by-step explanation:

6 0
3 years ago
What is the angle measure of an arc bounding sector with an area of 5pie square miles?
Snowcat [4.5K]

if the diameter is 20, the its radius must be half that or 10.

\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}

8 0
2 years ago
The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for t
nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

8 0
3 years ago
Find the surface area of the prism.
Katen [24]

Answer:

204

Step-by-step explanation:

6*10=60

6*6=36

6*8=48

2((10*6)/2)=60

60+60+36+48=204

8 0
2 years ago
Read 2 more answers
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