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3 years ago

Area of the desk is 1 4/5 square meters the length of the desk is 2 meters what is the width of the desk

1 answer:

5 0

Assuming that the desk simply has a shape of a rectangle, then the formula of the area is simply the product of length and width:

Area = length * width

So the width is:

width = Area / length

width = (9/5 m^2) / (2 m)

width = 0.9 meters

You might be interested in

The hardcover version of a book weighs 7 ounces while its paperback version weighs 5 ounces. Forty-five copies of the book weigh

konstantin123 [22]

Answer:

33 copies were paperback and 12 were hardcover.

Step-by-step explanation:

Let h represent the number of hardcover copies and p represent the number of paperback copies.

We know that the total number of copies was 45; this gives us the equation

h+p = 45

We know that each hardcover copy is 7 ounces; this gives us the expression 7h.

We also know that each paperback copy is 5 ounces; this gives us the expression 5p.

We know that the total weight was 249 ounces; this gives us the equation

7h+5p = 249

Together we have the system

$\left \{ {{h+p=45} \atop {7h+5p=249}} \right.$

We will use elimination to solve this. First we will make the coefficients of the variable p the same; to do this, we will multiply the top equation by 5:

$\left \{ {{5(h+p=45)} \atop {7h+5p=249}} \right. \\\\\left \{ {{5h+5p=225} \atop {7h+5p=249}} \right.$

To eliminate p, we will subtract the equations:

$\left \{ {{5h+5p=225} \atop {-(7h+5p=249)}} \right. \\\\-2h=-24$

Divide both sides by -2:

-2h/-2 = -24/-2

h = 12

There were 12 hardcover copies sold.

Substitute this into our first equation:

12+p=45

Subtract 12 from each side:

12+p-12 = 45-12

p = 33

There were 33 paperback copies sold.

3 0

3 years ago

Read 2 more answers

Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - α)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of t for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

6 0

3 years ago

For a normal distribution with μ=500 and σ=100, what is the minimum score necessary to be in the top 60% of the distribution?

STatiana [176]

Answer:

475.

Step-by-step explanation:

We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.

We will use z-score formula and normal distribution table to solve our given problem.

$z=\frac{x-\mu}{\sigma}$