Answer:
f(g(5)) = 11
Step-by-step explanation:
- f(x) = x - 2
- g(x) = 3x - 2
The composite function f(g(5)) is found by plugging g(5) for x into f(x).
Let's find g(5) first.
- g(5) = 3(5) - 2
- g(5) = 15 - 2
- g(5) = 13
Next, plug this value for x into f(x).
- f(13) = (13) - 2
- f(13) = 11
∴ <u>f(g(5)) = 11</u>.
247,039 rounded to the nearest thousand becomes
247,000
In this questions, there are a few functions with more/less than and equal to symbol.
1. Answer: f(x)= 3
f(x)= 2<x ≤3
The x is [more than 2] and [less than 3] or [equal to 3]. The number that will fulfill the condition would be only 3 since 3 is [more than 2] and [equal to 3].
2. f(x)= 1
f(x)= 0<x ≤1
The x is [more than 0] and [less than 1] or [equal to 1]. The number that will fulfill the condition would be only 1 since 1 is [more than 0] and [equal to 1].
3. f(x)= x
f(x)= 1≤x ≤2
The x is [more than 1] or [equal to 1] and [less than 2] or [equal to 2]. The number that will fulfill the condition would be
1, since 1 is [equal to 1] and [less than 2]
2, since 2 is [more than 1] and [equal to 2]
The graph that follow the value would be: f(x)= x
4. Answer: f(x)= 4
f(x)= 3<x ≤4
The x is [more than 3] and [less than 4] or [equal to 4]. The number that will fulfill the condition would be only 4 since 4 is [more than 3] and [equal to 4].
Answer
Find out the how many shorter pieces can a seamstress cut.
To prove
Let us assume that the number of shorter pieces cut be x.
As given
A seamstress has a piece of cloth that is 3 yards long.
she cuts it into shorter lengths of 16 inches each.
As
1 yards = 36 inches
Now convert 3 yards into inches.
= 3 × 36
= 108 yards
Than the equation becomes
16x = 108

x = 6.75
Therefore the number of shorter pieces be 6 .
Answer:
Step-by-step explanation:
Hello!
The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.
For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.
The established variable has a normal distribution with a standard deviation of σ= 2 min.
From the taken sample an average time difference of X[bar]= 0.77 was obtained.
If the app is correct, the true mean should be around cero, symbolically: μ=0
a. The hypotheses are:
H₀:μ=0
H₁:μ≠0
b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.
c.
![Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }](https://tex.z-dn.net/?f=Z_%7BH_0%7D%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)

d. This hypothesis test is two-tailed and so is the p-value.
p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058
e. 90% CI

X[bar] ± 
0.77 ± 1.645 * 
[0.098;1.442]
I hope this helps!