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2 years ago

Which inequality will have a shaded area above its graph? a. 2x − y ≥ 4 b. 4x + 3y < 6 c. x − 9y ≤ 1 d. x − 3y > 4

Mathematics

2 answers:

jekas [21]2 years ago

6 0

Answer:

c) x − 9y ≤ 1 have shaded area above of graph.

Step-by-step explanation:

Given : a. 2x − y ≥ 4

b. 4x + 3y < 6

c. x − 9y ≤ 1

d. x − 3y > 4

To find : Which inequality will have a shaded area above its graph.

Solution : x -9y ≤ 1

Now we will solve for y

On subtracting by x both side

-9y ≤ 1 -x

On dividing by -9 both sides

y ≥ $\frac{1-x}{-9}$ .

As we now that Shade above the line of graph for a "greater than" (y> or y≥)

Therefore, c) x − 9y ≤ 1 have shaded area above of graph.

Pavlova-9 [17]2 years ago

3 0

c. x − 9y ≤ 1 is the inequality having a shaded area above its graph.The line separating the shaded part will be a solid line as the inequality involves equals to sign.
See the attachment for the graph.

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To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assig

Keith_Richards [23]

Answer:

1. Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part 1

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

We need to find the mean for each group first and the grand mean.

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

If we apply the before formula we can find the mean for each group

$\bar X_A = 27$ , $\bar X_B = 24$ , $\bar X_C = 30$ . And the grand mean $\bar X = 27$

Now we can find the sum of squares between:

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

Each group have a sample size of 4 so then $n_j =4$

$SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72$

The degrees of freedom for the variation Between is given by $df_{between}=k-1=3-1=2$ , Where k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

$MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36$

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

$SS_{error}=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306$

And the degrees of freedom are given by:

$df_{within}=N-k =3*4 -3 = 12-3=9$ . N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

$MSE=\frac{SS_{within}}{N-k}=\frac{306}{9}=34$

C. 34

Part 4

The test statistic F is given by this formula:

$F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059$

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is $F_{crit}=8.02$

B. 8.02

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2 years ago

The moon is 384,403 km from the Earth. Estimate how many quarters laid end to end it would take to reach the moon if a quarter h

Kryger [21]

$16.96 16 times 6% is .96

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2 years ago

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A private plane is traveling due east at a rate of 160 mph. A south wind is blowing 30 mph. What is the actual velocity of the p

photoshop1234 [79]

The resultant velocity of the plane is the sum of the two velocity vectors which are perpendicular to each other. See the attached figure.

The magnitude of the resultant velocity is

$\sqrt{160^2+30^2}= \sqrt{26500}=162.79 \;mph$ .

The approximate value of the actual velocity of the plane is $163 \;mph$ . Correct choice is (D).

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2 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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1 year ago

A pebble is tossed into the air from the top of a cliff. The height, in feet, of the pebble over time is modeled by the equation

Charra [1.4K]

Answer:

Step-by-step explanation:

The maximum height reached by the pebble modeled by the quadratic function,

$h(t)=-16t^2+32t+80$ , can be found by finding the vertex.