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Marizza181 [45]
2 years ago
13

According to the text, approximately what percentage of college students drop out before obtaining a degree

Mathematics
1 answer:
BabaBlast [244]2 years ago
4 0

Answer:

40% but there is no article so I looked it up

Step-by-step explanation:

You might be interested in
What is the value of x?<br> 160°<br> x =<br> 519
forsale [732]

Answer:

x = 58

Step-by-step explanation:

Angle Formed by Two Secants = 1/2(difference of Intercepted Arcs)

51 =1/2 (160-x)

Multiply each side by 2

102 = 160-x

Subtract 160 from each side

102-160 = -x

-58 = -x

Multiply by -1

58 =x

6 0
3 years ago
Read 2 more answers
Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1
Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

P(x=r)=^nC_rp^rq^{n-r}

(a) Exactly two will be drunken drivers.

P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}

P(x=2)=66(0.2)^{2}(0.8)^{10}

P(x=2)=\approx 0.28347

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)

P(x=3\text{ or }x=4)=P(x=3)+P(x=4)

Using binomial we get

P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}

P(x=3\text{ or }x=4)=0.236223+0.132876

P(x=3\text{ or }x=4)\approx 0.369099

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

P(x\geq 7)=1-P(x

P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]

P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]

P(x\leq 7)=1-[0.9961]

P(x\leq 7)=0.0039

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)

P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315

P(x\leq 5)=0.9806

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0
3 years ago
2. The water level of a river was measured each day during a two-week period. The
vampirchik [111]

Step-by-step explanation:

day during a two-week period. The

graph models the linear relationship between the water level of the river in feet

and the number of days the water level was measured.

Water Level of River

The initi

28

Ο Ο Ο

The max

24

20

The wate

16

Water Level (ft)

12

o

The water

CLEAR ALL

4

0

2

4

12

6 8 10

Number of Days

Which statement best describes the y-intercept of the graph?

< PREVIOUS

O 3

Os Oo Or

0.8

HH

7 0
2 years ago
Read 2 more answers
Which of the numbers below is less than zero? Select all that apply.
UkoKoshka [18]

Answer:

A, B, E area ll less than 0

Step-by-step explanation:

They are all negative numbers.

4 0
2 years ago
Read 2 more answers
Me podrían ayudar y explicarme el proceso please
kakasveta [241]

Answer: Encontrarás las respuestas debajo de cada explicación. Espero que consideres darme brainliest y 5 estrellas, y más importante, que hayas entendido.

Step-by-step explanation:

Esto es la regla de 3. Conoces 3 valores y necesitas encontrar un 4to valor.

Ya sabemos que 1 caja pesa 20kg,

1 - 20kg

ahora queremos saber cuantas cajas (x) equivalen a 7653kg

x - 7653kg

Hacemos un pequeño cuadro poniendo en un lado los numeros de cajas y del otro el peso.

Pongamos la cantidad de cajas en la izquierda y el peso en la derecha

\frac{1}{x} =\frac{20kg}{7653kg}

Procedemos a multiplicar en cruz. Solo podemos multiplicar si tenemos ambos valores. En este caso, los valores en cruz que tenemos son 1 y 7653kg porque en la otra cruz (20 y x) tenemos nuestra incognita. Y, por ultimo, dividimos entre el numero cuya cruz es con la incognita (20kg)

Esto nos deja la expresión así;

x=\frac{1*7653kg}{20kg}

x=382.65

Como una caja no puede ser un numero decimal, redondeamos.

x=383

Por lo tanto, se necesitan aproximadamente 383 cajas para que su peso equivalga a 7653 kg.

-------------------------------------------------------------------------------------------------------

La otra pregunta dice: Si una caja pesa 20kg, (1 - 20kg), ¿Cuántas cajas se necesitan para equivaler 9500kg? (x - 9500kg)

Hacemos lo mismo que arriba, ponemos la cantidad de cajas de un lado, y el peso de otro lado.

\frac{1}{x}=\frac{20kg}{9500kg}

Multiplicamos aquellos valores en cruz que conozcamos y dividimos por el valor cuya cruz sea con x.

x=\frac{1*9500kg}{20kg}

x=475

Esta es la cantidad de cajas que se necesitan para que su pesa equivalga a 9500kg.

--------------------------------------------------------------------------------------------------------

Por último, la pregunta dice, ¿Cuántos kilogramos hay en 873 cajas?

Ya sabemos que una caja pesa 20kg ( 1 - 20kg ) y ahora buscamos cuantos kg hay en 873 cajas; es decir, (873 - x)

Nos quedaría así;

\frac{1}{873}=\frac{20kg}{x}

Ahora, nuestra cruz es 873 y 20kg, y el divisor es 1.

x=\frac{873*20kg}{1}

x=17460kg

Este sería el peso de 873 cajas.

5 0
3 years ago
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