Hm
10 + 4 x 2 - 3
=10 + 8 - 3
=18 - 3
=15
Hi there!
To solve this problem using substitution, we need to set x equal to a value and substitute it into the other equation.
Since
x - 4y = 5,
x = 5 + 4y
3x - 7y = 10
3(5 + 4y) - 7y = 10
15 + 12y - 7y = 10
15 + 5y = 10
5y = -5
y = -1
Now that we know y is -1, we can substitute it back into the equation to find x:
y = -1
x - 4y = 5
x - 4(-1) = 5
x - (-4) = 5
x + 4 = 5
x = 1
So, your answers are x = 1 and y = -1.
Hope this helps!
Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:
P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,
0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Answer:
Step-by-step explanation:
Part A
Part B
- 5.50x = 62.50 - 35
- 5.50x = 27.50
- x = 27.50/5.50
- x = 5
The appropriate measure of central tendency is one that shows
difference and is suitable for a scale that is nominal.
Response:
- The measure of central tendency to use is the <u>mode</u>.
<h3>How can the appropriate measure of central tendency be selected?</h3>
The mean is the sum of the measurements divided by the number count
of the plants.
The mode is the measurement that has the highest frequency.
The median is the measurement of the middle plant when arranged in a
given order according to size.
To argue that there is a difference between the plants, the measure of
central tendency to use is the mode, given that the data involves
measurements which can be expressed in a nominal scale.
Therefore;
- The measure of central tendency that will be best for Mrs. Hull to use is the<u> mode</u>
Learn more about the measures of central tendencies here:
brainly.com/question/1027437
