In a group of 36 children, there are twice as many boys as girls, how many girls are there?

ANSWER FOR BRAINLEST (BEST TO EXPLAIN WILL GET IT)

lina2011 [118]

Answer:

-7g+13

Step-by-step explanation:

b) because combining like terms results in:

g-8g=7g

and

15-2= 13

5 0

3 years ago

Let X be the time it takes for a person to choose a birthday gift, where X has an average value of 27 minutes. If the random var

jeka57 [31]

Answer:

The parameters of this exponential distribution is $\lambda$ = $\frac{1}{27}$ .

Step-by-step explanation:

We are given that the random variable X is known to be exponentially distributed and let X be the time it takes for a person to choose a birthday gift, where X has an average value of 27 minutes.

<u><em>So, X = time it takes for a person to choose a birthday gift</em></u>

The probability distribution function of exponential distribution is given by;

$f(x) = \lambda e^{-\lambda x} , x >0$ where, $\lambda$ = parameter of distribution.

Now, the mean of exponential distribution is = $\frac{1}{\lambda}$ which is given to us as average value of 27 minutes that means $\lambda = \frac{1}{27}$ .

So, X ~ Exp( $\lambda = \frac{1}{27}$ ) .

Therefore, the parameter of this exponential distribution is $\lambda$ .

5 0

4 years ago

Suppose the object moving is Dave, who hasamassofmo =66kgatrest. Whatis Daveâs mass at 90% of the speed of light? At 99% of the

nignag [31]

Step-by-step explanation:

Formula that relates the mass of an object at rest and its mass when it is moving at a speed v:

$m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}$

Where :

m = mass of the oebjct in motion

$m_o$ = mass of the object when at rest

v = velocity of a moving object

c = speed of the light = $3\times 0^8 m/s$

We have :

1) Mass of the Dave = $m_o=66 kg$

Velocity of Dave ,v= 90% of speed of light = 0.90c

Mass of the Dave when moving at 90% of the speed of light:

$m=\frac{66 kg}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}$

m = 151.41 kg

Mass of Dave when when moving at 90% of the speed of light is 151.41 kg.

2) Velocity of Dave ,v= 99% of speed of light = 0.99c

Mass of the Dave when moving at 99% of the speed of light:

$m=\frac{66 kg}{\sqrt{1-\frac{(0.99c)^2}{c^2}}}$

m = 467.86 kg

Mass of Dave when when moving at 99% of the speed of light is 467.86 kg.

3) Velocity of Dave ,v= 99.9% of speed of light = 0.999c

Mass of the Dave when moving at 99.9% of the speed of light:

$m=\frac{66 kg}{\sqrt{1-\frac{(0.999c)^2}{c^2}}}$

m = 1,467.17 kg

Mass of Dave when when moving at 99.9% of the speed of light is 1,467.17 kg.

4) Mass of the Dave = $m_o=66 kg$

Velocity of Dave,v=?

Mass of the Dave when moving at v speed of light: 500

$500 kg=\frac{66 kg}{\sqrt{1-\frac{(v)^2}{(3\times 10^8 m/s)^2}}}$

$v=2.973\times 10^8 m/s$

Dave should be moving at speed of $2.973\times 10^8 m/s$ .

4 0

3 years ago

A fair coin is tossed and a single number cube is rolled. the probability of rolling a number less than 5 is ________________. t

Alik [6]

You have a 4/ 6 chance
then you have a 2/6 chance if the number is an odd number then reducing you would get
1. 2/3
2. 1/3

7 0

3 years ago

Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 104 eligible vot

crimeas [40]

Answer:

The probability that exactly 27 of 104 eligible voters voted is 0.057 = 5.7%.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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 zscore 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 pvalue, 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this case, assume that 104 eligible voters aged 18-24 are randomly selected.

This means that $n = 104$ .

Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

This means that $p = 0.22$

Mean and standard deviation:

$\mu = 104*0.22 = 22.88$

$\sigma = \sqrt{104*0.22*0.78} = 4.2245$

Probability that exactly 27 voted

By continuity continuity, 27 consists of values between 26.5 and 27.5, which means that this probability is the p-value of Z when X = 27.5 subtracted by the p-value of Z when X = 26.5.

X = 27.5

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

$Z = \frac{27.5 - 22.88}{4.2245}$