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2 years ago

Jack bought a shirt for $14 because it was discounted 30%. what is the regular price?

Mathematics

1 answer:

Andru [333]2 years ago

8 0

All you have to do is 14*.30 which equals to $4.20

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Look at the picture and help answer question 7 please

qwelly [4]

3pm we know this because 5^3= 125 so then we plug in numbers for the exponent in this case 6 which equals 15,625 now 9am plus 6 is 3pm

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2 years ago

Find the coordinates of a point that divides the directed line segment PQ in the ratio 5:3. A) (2, 2) B) (4, 1) C) (–6, 6) D) (4

Yuri [45]

Answer:

The answer is explained below

Step-by-step explanation:

The question is not complete we need point P and point Q.

let us assume P is at (3,1) and Q is at (-2,4)

To find the coordinate of the point that divides a line segment PQ with point P at $(x_1,y_1)$ and point Q at $(x_2,y_2)$ in the proportion a:b, we use the formula:

$x-coordinate:\\\frac{a}{a+b}(x_2-x_1)+x_1 \\\\While \ for\ y-coordinate:\\\frac{a}{a+b}(y_2-y_1)+y_1$

line segment PQ is divided in the ratio 5:3 let us assume P is at (3,1) and Q is at (-2,4). Therefore:

$x-coordinate:\\\frac{5}{5+3}(-2-3)+3 \\\\While \ for\ y-coordinate:\\\frac{5}{5+3}(4-1)+1$

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2 years ago

There is a book sale at the libaray the price for each book is $4 which expression can be used to show how much money the librar

ASHA 777 [7]

4 x 289= the total cost

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2 years ago

Identify the pattern then write the next three terms in each sequence 192,96,48,24,...

Karo-lina-s [1.5K]

1st. Find the pattern
Pattern is dividing by 2.
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12/2=6
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2 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

2 years ago