???? Help me with my work

Need answer quick please answer quick

Natalka [10]

Answer : 1⅖ft.²

Given :-

Print is 7/8 ft wide and 1⅗ ft long .

To Find :-

The area of the print .

Here the length of the print is 1⅗ ft . So on converting this into improper fraction , we have ;

→ length = 1⅗ ft = (5*1+3)/5 ft = (5+3)/5ft = 8/5ft.

Also ,

→ breadth = 7/8ft .

We know that the area of a rectangle is ,

→ Area = length * breadth

On substituting the respective values,

→ Area = 8/5 ft. * 7/8ft

→ Area = 7/5 ft.²

After converting improper fraction into mixed fraction ,

→ Area = 1⅖ ft² .

Henceforth the area of the print is 1⅖ft.² .

I hope this helps.

6 0

3 years ago

The graph of y > 1 2 x - 2 is shown. Which set contains only points that satisfy the inequality? A) {(6, 1), (-1, -3), (4, 4)

Kaylis [27]

Answer:

Option B) {(1, -1), (-3, -3), (2, 4)}

Step-by-step explanation:

The correct inequality is

$y>\frac{1}{2}x-2$

The solution of the inequality is the shaded area above the dashed line $y=\frac{1}{2}x-2$

see the attached figure to better understand the problem

we know that

If a ordered pair satisfy the inequality, then the ordered pair must lie in the shaded area of the solution

Verify each case

case A) {(6, 1), (-1, -3), (4, 4)}

ordered pair (6,1)

For x=6, y=1

substitute in the inequality

$1>\frac{1}{2}(6)-2$

$1>1$ ---> is not true

therefore

The point not satisfy the inequality

case B) {(1, -1), (-3, -3), (2, 4)}

ordered pair (1,-1)

For x=1, y=-1

substitute in the inequality

$-1>\frac{1}{2}(1)-2$

$-1>-1.5$ ---> is true

so

The point satisfy the inequality

ordered pair (-3,3)

For x=-3, y=3

substitute in the inequality

$3>\frac{1}{2}(-3)-2$

$3>-3.5$ ---> is true

so

The point satisfy the inequality

ordered pair (2,4)

For x=2, y=4

substitute in the inequality

$4>\frac{1}{2}(2)-2$

$4>-1$ ---> is true

so

The point satisfy the inequality

therefore

The set contains only points that satisfy the inequality

case C) {(1, -1), (-3, -3), (4, -2)}

ordered pair (4,-2)

For x=4, y=-2

substitute in the inequality

$-2>\frac{1}{2}(4)-2$

$-2>0$ ---> is not true

therefore

The point not satisfy the inequality

case D) {(-1, -3), (-3, -3), (2, 4)}

ordered pair (-1,-3)

For x=-1, y=-3

substitute in the inequality

$-3>\frac{1}{2}(-1)-2$

$-3>-2.5$ ---> is not true

therefore

The point not satisfy the inequality

4 0

3 years ago

The sum of two numbers is 27 and the difference is 19. Find the 2 numbers.

zhuklara [117]

Answer:

the sum of two numbers is 27. their difference is 5. find the number

Step-by-step explanation:

Let x be the first number and y be the second number ... so we have

x + y = 27

x - y = 5 Adding the equations, we have

2x = 32 Dividing through by 2, we have

x = 16 And substituting or x in the first equation, we have

16 + y = 27 Subtract 16 from both sides

y = 11

4 0

4 years ago

The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit

Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have 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 $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$