198.9 grams equals how many ounces?

The miles-per-gallon obtained by the 1995 model Z cars is normally distributed with a mean of 22 miles-per-gallon and a standard

AlekseyPX

Answer:

0.42074

Step-by-step explanation:

In this question, we are asked to calculate the probability that a car will get less than 21 miles per gallon.

This is P(x < 21)

z = 21-22/5 = -1/5 = -0.2

Therefore the probability of getting less than 21 miles per gallon will be;

P(x<21) = P (z < -0.2)

= 0.42074

4 0

3 years ago

Solve the system of equations by substitution. 3/8x + 1/3y = 17/24 x + 7y = 8

bixtya [17]

Using the second equation, let’s find x

x + 7y = 8
x = 8 - 7y

Substitute x in the other equation

3/8(8 - 7y) + 1/3y = 17/24
3 - 21/8y + 1/3y = 17/24
-55/24y = -55/24
y = 1

Plug in y in any equation

x + 7(1) = 8
x + 7 = 8
x = 1

x = 1 and y = 1

3 0

3 years ago

Which point satisfies the inquality 2×+y > 10 a. (2,3) b.(3,4) c. (3,2) d. (4,3)

Anna11 [10]

D. (4,3) that's the answer

8 0

4 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago

A parabola has a focus of F(2, -0.5) and a directrix of y=-1.5 P(x,y) represents any point on the parabola, while D(x, -1.5) rep

prohojiy [21]

The sketch of the parabola is attached below

We have the focus $(a,b) = (2, -0.5)$
The point $P(x,y)$
The directrix, c at $y=-1.5$

The steps to find the equation of the parabola are as follows

Step 1
Find the distance between the focus and the point P using Pythagoras. We have two coordinates; $(2, -0.5)$ and $(x,y)$ .
We need the vertical and horizontal distances to find the hypotenuse (the diagram is shown in the second diagram).
The distance between the focus and point P is given by
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$

Step 2
Find the distance between the point P to the directrix $c$ . It is a vertical distance between y and c, expressed as $y-c$

Step 3
The equation of parabola is then given as
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$ = $y-c$
$(x-a)^{2}+ (y-b)^{2}= (y-c)^{2}$ ⇒ substituting a, b and c
$(x-2)^{2}+ (y--0.5)^{2} = (y--1.5)^{2}$
$(x-2)^{2}+ (y+0.5)^{2}= (y+1.5)^{2}$ ⇒Rearranging and making $y$ the subject gives

$y= \frac{ x^{2} }{2} -2x+1$

7 0

4 years ago