All categories

2 years ago

What percent is equivalent to 24/50? 24% 48% 50% 52%

Mathematics

2 answers:

stealth61 [152]2 years ago

8 0

Answer:

48%

Step-by-step explanation:

24 is almost 25 which is half of 50 you it would be 48%, hope this helps!:)

Dmitry [639]2 years ago

3 0

48 percent because 24/50 is almost 25 which would make it 50 put it’s not so it’s 48 have a great day

You might be interested in

How do you write an equation in standard form given point (2,2) (0,-4)

notka56 [123]

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (2, 2) and (0, -4). Substitute:

$m=\dfrac{-4-2}{0-2}=\dfrac{-6}{-2}=3\\\\y-2=3(x-2)$

The standard form:

$Ax+By=C$

$y-2=3(x-2)$ <em>use distributive property</em>

$y-2=3x-6$ <em>add 6 to both sides</em>

$y+4=3x$ <em>subtract y from both sides</em>

$4=3x-y$

<h3>Answer: 3x - y = 4</h3>

6 0

3 years ago

Which statement is correct? A) 13/ 3 < 4.25 < 43 /10 B) 43 /10 < 13 /3 < 4.25 C) 4.25 < 13 /3 < 43/ 10 D) 4.25

Ivan

D. 4.25 < 43/10 < 13/3

4.25 = 4.25
43/10 = 4.3
13/3 = 4.33

4.25 < 4.3 < 4.33

4 0

3 years ago

Read 2 more answers

Multiply.

icang [17]

Sorry for the delay but I answered your question and the answer is -6xsquare +11x-4

4 0

2 years ago

Read 2 more answers

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

3 years ago

Find the area of a trapezoid with bases that measure 9.2 CM and 7.7 cm and height 6.8 cm

maria [59]

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\
---------\\
a=9.2\\
b=7.7\\
h=6.8
\end{cases}\implies A=\cfrac{6.8(9.2+7.7)}{2}$

8 0

3 years ago

Read 2 more answers