The perimeter of a rectangle is 343434 units. Its width is 6.56.56, point, 5 units.

Mathematics

2 answers:

Greeley [361]2 years ago

5 0

Answer:

the equation is 2l + 13 = 34 then l is 10.5

Step-by-step explanation:

Andru [333]2 years ago

3 0

This is annoying

the perimiter is 34 units
the width is 6.5 units

ok.
perimiter=2(Length+Width)
P=2(L+W)
solve for L
distribute
P=2L+2W
minus 2W
P-2W=2L
divide by 2
$\frac{P-2W}{2}=L$

given that P=34 units and W=6.5 units
$\frac{34-2(6.5)}{2}=L$

the equation would be $L=\frac{P-2W}{2}$ or $L=\frac{P}{2}-W$

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Helena needs 3.5 cups of flour per loaf of bread and 2.5 cups of flour per batch of muffins. She also needs 0.75 cup of sugar pe

Arlecino [84]

From the given data, we can generate two equations with two unknowns.

We let x = number of loaves of bread
y = number of batches of muffins

For the equation of the flour requirement:
17 = 3.5x + 2.5y

For the equation of the sugar requirement:
4.5 = 0.75x + 0.75y

We evaluate the solutions by manipulating one of the equations into terms of the other. We use the first equation.We write x in terms of y.

x = (4.5/0.75) - y

Substitute the third equation to the second equation.

17 = (3.5((4.5/0.75)-y)) + 2.5y

Evaluating y and x, we have,

y = 4 and x = 2

Thus, from the amounts she has in hand, she can make 4 loaves of bread and 2 batches of muffins.

4 0

2 years ago

Read 2 more answers

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1

BigorU [14]

Answer:

a: Sample mean: 20, Sample standard deviation: 1.732

b: 19.13 <µ < 20.87

c: 18.84< µ < 21.16

d: as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a: Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

The t-value is 1.415

See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

The t-value is 1.895

See the third attached photo for the construction of the confidence interval