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leva [86]
3 years ago
15

Cost of a dozen pens is ₹ 180 and cost of 8 ball pens is ₹ 56. Find the ratio of the

Mathematics
1 answer:
kirza4 [7]3 years ago
5 0

Answer:

Ratio of cost of a pen to cost of a ball pen is 15:7

Step-by-step explanation:

Given that:

Cost of dozen pens = ₹180

Unit rate = \frac{180}{12}

Unit rate = ₹ 15 per pen

Cost of 8 ball pens = ₹ 56

Unit rate = \frac{56}{8}

Unit rate = ₹7 per ball pen

Ratio of cost of a pen to cost of a ball pen,

15 : 7

Hence,

Ratio of cost of a pen to cost of a ball pen is 15:7

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PLEASE HELP!!!!
Bogdan [553]

Answer:

Step-by-step explanation:

We would apply the formula for binomial distribution which is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - r) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 18% = 18/100 = 0.18

q = 1 - p = 1 - 0.18

q = 0.82

n = 5

Therefore,

P(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

P(x = 0) = 5C0 × 0.18^0 × 0.82^(5 - 0)

P(x = 0) = 0.37

P(x = 1) = 5C1 × 0.18^1 × 0.82^(5 - 1)

P(x = 1) = 0.41

P(x = 2) = 5C2 × 0.18^2 × 0.82^(5 - 2)

P(x = 2) = 0.18

Therefore,

P(x ≤ 2) = 0.37 + 0.41 + 0.18 = 0.96

5 0
2 years ago
5. Put the following numbers in order from least to greatest in the spaces provided.
zloy xaker [14]

Answer:

7,7.5,8,8.1

Step-by-step explanation:

count 1 to 10. See what comes first and that's how you determine what comes first.

5 0
3 years ago
The ground temperature at an airport is 16 °C. The temperature decreases by 5.6 °C for every increase of 1 kilometer above the g
love history [14]

0 km = 16

1km = 16 - 5,6 = 10,4

2km = 10,4 - 5,6 = 4,8

3km = 4,8 - 5,6 = -0,8

4km = -0,8 - 5,6 = -6,4

5km = -6,4 - 5,6 = -12

3 0
2 years ago
Which is the value of this expression when j = negative 2 and k = negative 1?
Natalija [7]

Answer:

Option A.

Step-by-step explanation:

It is given that

j=-2, k=-1

The given expression is

\left(\dfrac{jk^{-2}}{j^{-1}k^{-3}}\right)^3

Substitute the given values in the above expression.

\left(\dfrac{(-2)(-1)^{-2}}{(-2)^{-1}(-1)^{-3}}\right)^3

\Rightarrow \left(\dfrac{(-2)(1)}{\dfrac{1}{-2}(-1)}\right)^3

\Rightarrow \left(\dfrac{-2}{\dfrac{1}{2}}\right)^3

\Rightarrow \left(-4\right)^3

\Rightarrow -64

Therefore, the correct option is A.

3 0
3 years ago
Read 2 more answers
Using the technique in the model above, find the missing sides in this 30°-60°-90° triangle.
lina2011 [118]
For this case what you should do is use the following trigonometric relationship:
 sin (x) = C.O / h
 Where
 x: angle
 C.O: opposite leg
 h: hypotenuse
 Substituting the values we have:
 sen (60) = long / h
 sen (60) = 3 / h
 h = 3 / sin (60)
 h = 3.46
 Answer:
 h = 3.46
7 0
3 years ago
Read 2 more answers
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