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xeze [42]
2 years ago
7

2/5 (s + 20) pls help

Mathematics
1 answer:
nevsk [136]2 years ago
6 0

Answer:

2s/5 + 8

Step-by-step explanation:

if you wanted it simplified..

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.0003 is 1/10?of which decimal
jarptica [38.1K]
0.0003 would be 1/10 of 0.00003. If you divided 0.00003 by 10 you would get 0.0003. 
Hopefully, this was useful.
4 0
3 years ago
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The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.
Monica [59]

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = \frac{b}{a}

- This is a vertical hyperbola, the equation is

       \frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

       \frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1

       \frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1

7 0
3 years ago
Read 2 more answers
9.4 The heights of a random sample of 50 college stu- dents showed a mean of 174.5 centimeters and a stan- dard deviation of 6.9
gladu [14]

Answer:  a) (176.76,172.24), b) 0.976.

Step-by-step explanation:

Since we have given that

Mean height  = 174.5 cm

Standard deviation = 6.9 cm

n = 50

we need to find the 98% confidence interval.

So, z = 2.326

(a) Construct a 98% confidence interval for the mean height of all college students.

x\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=(174.5\pm 2.326\times \dfrac{6.9}{\sqrt{50}})\\\\=(174.5+2.26,174.5-2.26)\\\\=(176.76,172.24)

(b) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centime- ters?

Error would be

\dfrac{\sigma}{\sqrt{n}}\\\\=\dfrac{6.9}{\sqrt{50}}\\\\=0.976

Hence, a) (176.76,172.24), b) 0.976.

8 0
3 years ago
Of the 58 employees in a firm, 8 are considered support staff, and 7 of the employees have more than five years of experience. A
polet [3.4K]

Using statistical concepts, it is found that the number of outcomes that are possible for the complement of the union of Events J and K is of 43.

<h3>What is the union of events J and K?</h3>

It means that at least one of event J or event K is true, hence, it is composed by employees that are either considered support staff(less than 5 years of experience) or employees that have more than five years of experience, combining a total of 7 + 8 = 15 employees.

<h3>What is the complement?</h3>

The total number of outcomes of the union of J and K, plus the complement, add to the total number of 58, hence:

15 + x = 58

x = 43.

The number of outcomes that are possible for the complement of the union of Events J and K is of 43.

More can be learned about complementary events at brainly.com/question/9752956

6 0
1 year ago
What is the cost of one cranberry and one banana nut muffin?
Bezzdna [24]
So for the cranberry muffin, for every 12 (a dozen) it's $3
Banana nut muffin: for every 12, it's $4.32
take the $3 and divide it by 12 which is $0.25 for every cranberry muffin
take $4.32 divide by 12 and you get $0.36 for every banana nut muffin
so $0.25 + $0.36 = $0.61
Therefore the answer is B
8 0
2 years ago
Read 2 more answers
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