A car dealership had $114,000 in average monthly sales. If you made 30% of all the sales, what is your total sales? help me

A prize wheel at a county fair is divided into 9 equal sections. Each section on the wheel is marked with a letter. Each letter

Alik [6]

Your answer would be 9/27 but can be converted to 1/3. Hope this helps

8 0

3 years ago

A thread is being pulled off a spool at the rate of 40 cm per second. Find the radius of the spool

murzikaleks [220]

Answer:

r = 0.0424 cm (3sf)

Step-by-step explanation:

it makes 150 rev/sec

1 revolution = 1 circumference = 2pi*r

so every second the spool goes through 2*pi*r*150 of distance

we know that the rate of pulling is 40cm/sec

so 2pi*r*150 = 40

r = 40/(2pi*150), which is 0.04244131816 cm.

check: Circumference is 2pi*r = 0.2666

now multiply by 150rev/sec = 40 cm/sec

3 0

2 years ago

Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How

HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

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 interval.

Parameter of 5.2 per square yard:

This means that $\mu = 5.2r$ , in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

$P(X \geq 1) = 1 - P(X = 0) = 0.99$

Thus:

$P(X = 0) = 1 - 0.99 = 0.01$

We have that:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}$

Then

$e^{-5.2r} = 0.01$

$\ln{e^{-5.2r}} = \ln{0.01}$

$-5.2r = \ln{0.01}$

$r = -\frac{\ln{0.01}}{5.2}$

$r = 0.89$

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0

3 years ago

Sara hina and Arslan have a total of RS 390in their wallet . Sara has Rs. 7more than Hina .Arslan has 3 times what sara has . Ho

Brums [2.3K]

Answer:Sara hina and Arslan Have RS79.4,RS 72.4 and RS238.2 respectively.

Step-by-step explanation:

Step 1

Let the amount that hina has be x

the amount that sara has be represented as 7+x

and the amount that Arslan have be represented as 3(7+x)

such that the total amount in their wallet which is 390 can be expressed as

x+7+x + 3(7+x)=390

Step 2

Solving

x+7+x +21+3x=390

5x+28=390

5x==390-28

x=362/5=72.4

Hina has RS 72.4

Sara =7+x==72.4+77= RS 79.4

Arslan =3(7+x)=3 x 79.4=RS 238.2

5 0

3 years ago

A random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength

irinina [24]

Answer:

Probability that the sample mean comprehensive strength exceeds 4985 psi is 0.99999.

Step-by-step explanation:

We are given that a random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi.

Let $\bar X$ = sample mean comprehensive strength

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean comprehensive strength = 5500 psi

$\sigma$ = standard deviation = 100 psi

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample mean comprehensive strength exceeds 4985 psi is given by = P( $\bar X$ > 4985 psi)

P( $\bar X$ > 4985 psi) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{4985-5500}{\frac{100}{\sqrt{9} } }$ ) = P(Z > -15.45) = P(Z < 15.45)

= 0.99999

Since in the z table the highest critical value of x for which a probability area is given is x = 4.40 which is 0.99999, so we assume that our required probability will be equal to 0.99999.

4 0

3 years ago