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3 years ago

5

Find the solution set of each inequality below and then determine which inequalities have the same solution set as 1/3(-5x-3) &l

t;14

1 answer:

mario62 [17]3 years ago

5 0

1/3(-5x - 3) < 14

-5/3 x - 1 < 14

-5/3 x < 15

-5 x < 45

-x < 9

x > 9

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If testing the claim that sigma subscript 1 superscript 2 baseline not equals sigma subscript 2 superscript 2σ21≠σ22, what do w

12345 [234]

Answer:

The statistic for this system of hypothesis is given by:

$F=\frac{s^2_1}{s^2_2}$

If the statistic is equal to 1 then that means $s^2_1 = s^2_2$ and we don't have enough evidence to conclude that the two population variances and deviations are different.

Step-by-step explanation:

System of hypothesis

We want to test if the variation for a group1 is equal to another one 2, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \neq \sigma^2_2$

Calculate the statistic

The statistic for this system of hypothesis is given by:

$F=\frac{s^2_1}{s^2_2}$

If the statistic is equal to 1 then that means $s^2_1 = s^2_2$ and we don't have enough evidence to conclude that the two population variances and deviations are different.

6 0

3 years ago

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

3 0

2 years ago

The work of a student to solve a set of equations is shown below: m=8+2n; 4m=4+4n. Solve by elimination

nadya68 [22]

Answer:

Step-by-step explanation:

4(8+2n) = 4 + 4n

32 + 8n = 4+4n

4n = -28

n = -7

4 0

2 years ago

A city known for its temperature extremes started the day at -5 degrees Fahrenheit. The temperature increased by 78 degrees Fahr

charle [14.2K]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Starting temperature = - 5°F

Temperature at midday =increase in temperature by 78°F = (starting temperature + 78°F)

Temperature at nightfall = decrease in temperature by 32°F = ( temperature at midday - 32°F)

Temperature at midday:

(-5°F + 78°F) = 73°F

Temperature at nightfall:

(Temperature at midday - 32°F)

(73°F - 32°F) = 41°F

7 0

3 years ago

(20pts) the monthly average new york city temperature in fahrenheit are given below. Mont h jan feb mar apr may jun jul aug sep

HACTEHA [7]

Mean of the low is approximately 47.7 Median of the low is 45 and 50 or 47.5

3 0

2 years ago