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3 years ago

If 4^x+1 = 64, then what is the value of x^4

1 answer:

7 0

$4^{x} + 1 = 64 \\4^{x} = 63 \\ln(4^{x}) = ln(63) \\xln(4) = ln(63) \\x = \frac{ln(63)}{ln(4)}$

$(\frac{ln(63)}{ln(4)})^{4} = 79.78006702$

The value of x⁴ is equal to 79.8006702

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Calculamos el volumen (V) de un prisma cuadrangular expresado en unidades cúbicas. Respondemos: ¿cómo se calcula?...............

Vadim26 [7]

Answer:

a) El procedimiento queda resumido en lo que sigue:

1) Se halla el área de la base del cuadrilátero.

2) Se multiplica el resultado anterior por la longitud del prisma.

b) La expresión para calcular el volumen del prisma ( $V$ ), en unidades cúbicas, es:

$V = w\cdot h \cdot l$

Donde:

$w$ - Ancho de la base, en unidades.

$h$ - Altura de la base, en unidades.

$l$ - Longitud del prisma, en unidades.

Step-by-step explanation:

a) Dimensionalmente hablando, la unidad de volumen es igual al cubo de la unidad de longitud. Se explica a continuación los pasos necesarios para el cálculo de prisma cuadrangular:

1) Se halla el área de la base del cuadrilátero.

2) Se multiplica el resultado anterior por la longitud del prisma.

b) En consecuencia, se deriva la expresión para el volumen del prisma:

$V = A\cdot l$ (1)

Donde:

$A$ - Área de la base del prisma, en unidades cuadradas.

$l$ - Longitud del prisma, en unidades.

$V = w\cdot h \cdot l$ (2)

Donde:

$w$ - Ancho de la base, en unidades.

$h$ - Altura de la base, en unidades.

5 0

3 years ago

Write the next three terms of the arithmetic sequence.<br> First term: 2<br> Common difference: 13

IRINA_888 [86]

15, 28, 41.

add the common difference to each consecutive term to find the next term in the sequence.

2+ 13= 15
15+ 13= 28
28+ 13= 41

4 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Solve the equation: (x − 4)(x − 3) = 0

trasher [3.6K]

Answer:

x=4, x=3

Step-by-step explanation:

This is most often the format taken from the polynomial form of this problem, which is x^2-7x+12, and when it is in this format you just take each set of numbers within each parenthesis and set it to zero, then solve. In this case, x can be equal to either 3 or 4.

8 0

3 years ago

How many integers between 360 and 630 are there such that they have odd number of divisors?

Kryger [21]

Answer:

Step-by-step explanation:

Concept to Know: "A number is a perfect square if and only if it has odd number of positive divisors "

Find all the squared values that lies between 360 and 630

360< 19², 20², 21², 22², 23², 24², 25² < 630

19², 20², 21², 22², 23², 24², and 25² are all the squared values that lies between 360 and 630. There are seven of those squared numbers so the answer is 7.

6 0

3 years ago