Multiply. 35 22 x___

Use a scientific calculator to evaluate log 32.5 (round to the nearest hundredth).

AVprozaik [17]

Answer:

Log(32.5) ≈ 1.51

Step-by-step explanation:

*Look at the picture*

4 0

3 years ago

The points (2,4) and (-2,-4) are plotted on a coordinate plane the equation is y = a * x what is A?

laila [671]

A equals to 2.

Step-by-step explanation:

y = mx + c is the equation of the line and m is the slope and c is the y- intercept.

In the given equation, y = ax , a is the slope.

Slope A can be found by means of the formula, and using the points (2,4) and (-2-4)

slope, A = (y₂ -y₁) / (x₂-x₁)

Plugin the values as,

A = (-4-4) / (-2-2) = (-8) / (-4) = 2

So the value of A is 2.

5 0

2 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

Evaluate the summation of 3 times negative 2 to the n minus 1 power, from n equals 1 to 5. A) −93. B) −33. C) 33. D) 93

lions [1.4K]

∑ ( from n = 1 to n =5 ) 3 · ( -2 ) ^( n -1 )
a 1 = 3 · ( -2 ) ^0 = 3
a 2 = 3 · ( - 2 ) = - 6
a 3 = 3 · 4 = 12
a 4 = 3 · ( - 8 ) = - 24
a 5 = 3 · 16 = 48
3 - 6 + 12 - 24 + 48 = 33
Answer: C ) 33

4 0

2 years ago

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Multiply. 35 22 x____