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For each of the functions below, indicate whether the function is onto, one-to-one, neither orboth. If the function is not onto

Karo-lina-s [1.5K]

Answer:

$f_{1} :Z \rightarrow Z$ is one to one mapping, it is not onto mapping

Step-by-step explanation:

$f_{1} :Z \rightarrow Z\\ f_{1} (x) = x^{3}$

f₁(x) is one to one mapping

Let $x, y \epsilon Z$

f₁(x) = f₁(y):

x₁³ = y₁³

f₁(x) is not onto mapping

Example: If f₁(x) = 7,

x₁³ = 7

$x_{1} = \sqrt[3]{7}$

x₁ is not an element of Z

$f_{1} :Z \rightarrow Z$ is one to one mapping, it is not onto mapping

5 0

3 years ago

A surveyor, Toby, measures the distance between two landmarks and the point where he stands. He also measured the angles between

vagabundo [1.1K]

What we know so far:
Side 1 = 55m
Side 2 = 65m
Angle 1 = 40°
Angle 2 = 30°

What we are looking for:
Toby's Angle = ?
The distance x = ?

We need to look for Toby's angle so that we can solve for the distance x by assuming that the whole figure is a SAS (Side Angle Side) triangle.

Solving for Toby's Angle:
We know for a fact that the sum of all the angles of a triangle is 180°; therefore,
180° - (Side 1 + Side 2) = Toby's Angle
Toby's Angle = 180° - (40° + 30°)
Toby's Angle = 110°

Since we already have Toby's angle, we can now solve for the distance x by using the law of cosines r² = p²+ q²<span>− 2pq cos R where r is x, p is Side1, q is Side2, and R is Toby's Angle.
</span>
x² = Side1² + Side2² - 2[(Side1)(Side2)] cos(Toby's Angle)
x² = 55² + 65² - 2[(55)(65)] cos(110°)
x² = 3025 + 4225 -7150[cos(110°)]
x² = 7250 - 2445.44
x = √4804.56
x = 69.31m

∴The distance, x, between two landmarks is 69.31m

8 0

3 years ago

What are the factors of the mathematical expression 24p-27p²

Vlad [161]

Answer:

1expn=3p

2expn=8-9p

3p(8-9p)

6 0

3 years ago

Read 2 more answers

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$