Answer pls is mathematically is due today in 1 m

UUIVIY U

nasty-shy [4]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$2x -1 > x + 2$

Solve for x

Subtract x both sides

$2x -1-x > x + 2-x$

$x-1 > 2$

Adds 1 both sides

$x-1 +1> 2+1$

$x> 3$

The solution is the interval -------> (3,∞)

All real numbers greater than 3

In a number line is the shaded area at right of x=3 (open circle)

see the attached figure

7 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

2 years ago

For this question you will need to analyze data given in file seed Emergence. For your convenience, the data is posted as a text

Mrac [35]

Answer:

A) You will get the Mean, Sample variance and standard deviation, Minimum, maximum, range, and Boxplot.

B) The two-way ANOVA (analysis of variance) model is used for the two factors.

C) Outputs are obtained

D) Here, P-value (= 0.0377) < α (= 0.05). There is evidence that the treatments differ with respect to emerging plants.

Step-by-step explanation:

(A)

Use the MegaStat add-in in Excel to draw the boxplot and to find; the group counts, means, standard deviations, variances, minimums, maximums, and ranges for the 5 treatment groups.

In another Excel worksheet. Enter the data in 5 different columns, each column representing a treatment, and the first row holding the treatment names.

Go to Add-Ins > MegaStat > Descriptive Statistics.

Enter Sheet1!$A$1:$E$5 in Input range.

Tick on Mean, Sample variance and standard deviation, Minimum, maximum, range, and Boxplot.

Click OK.

(B)

Since there are two factors affecting the outcomes- the five treatments (Control, Arasan, Spergon, Semesan, and Fermate), and the four blocks, the two-way ANOVA (analysis of variance) model must be used.

(C)

We have used the Data Analysis tool-pack in Excel to construct the analysis of variance table.

We have arranged the data and entered it as follows:

Control, Arasan, Spergon, Semesan, Fermate

Block 1: 86 98 96 97 91

Block 2: 90 94 90 95 93

Block 3: 88 93 91 91 95

Block 4: 87 89 92 92 95

Go to Data > Data Analysis > Anova: Two-Factor Without Replication > OK.

In Input Range, enter $A$1:$E$6; tick on labels, enter Alpha as 0.05, and click OK.

The following output is obtained. Note that the analysis of variance table is given under “ANOVA” in the output.

(D)

In the analysis of variance table above, the “Rows” under “Sources of Variation” correspond to the treatments (as the observations under each treatment are noted along a row), whereas the “Columns” title relates to the block effects.

The P-value for Rows, hat is, for the treatments is 0.0377 (4 decimal places).

The level of significance is given as α = 0.05.

In this case, the null hypothesis is that, there is no significant difference between the 5 treatments, and the alternative hypothesis should be that, not all the treatments have the same effect.

The rejection rule for a test using the P-value is: Reject the null hypothesis, if P-value ≤ α. Otherwise, fail to reject the null hypothesis.

5 0

2 years ago

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Julli [10]

Lol hiiiiiiiiiiiiiiiiii

7 0

2 years ago

Which answer choice shows 21.97 written in expanded form? A. 2 + 1 + 9 + 7 B. 20 + 1 + 0.9 + 0.7 C. 20 + 10 + 0.9 + 0.07 D. 20 +

Simora [160]

Answer:

20 + 1 + 0.9 + 0.07

Step-by-step explanation:

Given the number 21.97 ;

To wite in expanded form:

Tens ___ Unit ____ tenth ____ hundredth

2_______ 1 _______ 9_________ 7

2 = 20

1 = 1

9 = 0.9

7 = 0.07

Hence,

20 + 1 + 0.9 + 0.07

6 0

3 years ago