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Natalija [7]
3 years ago
5

The area of a rectangular garden if 6045 ft2. If the length of the garden is 93 feet, what is its width?

Mathematics
2 answers:
AnnyKZ [126]3 years ago
4 0

Answer:

65 ft

Step-by-step explanation:

The area of a rectangle is

A = lw

6045 = 93*w

Divide each side by 93

6045/93 = 93w/93

65 =w

inna [77]3 years ago
4 0

Answer:

\huge \boxed{\mathrm{65 \ feet}}

Step-by-step explanation:

The area of a rectangle formula is given as,

\mathrm{area = length \times width}

The area and length are given.

6045=93 \times w

Solve for w.

Divide both sides by 93.

65=w

The width of the rectangular garden is 65 feet.

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Answer:

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Step-by-step explanation:

Sin B = opposite side/ hypotenuse = AC/AB

3 0
3 years ago
What does negative 4 over 7 &gt; −2 indicate about the positions of negative 4 over 7 and −2 on the number line?
mezya [45]

Smaller numbers sits on the left of greater numbers on the number line.

So, since

-\dfrac{4}{7} \geq -2

we're stating that -4/7 is larger and -2 is smaller. So, -4/7 sits on the right of -2 on the number line.

4 0
3 years ago
Read 2 more answers
For what values of x is the expression below defined?
AleksandrR [38]

The answer is x>0 on apex for x value

4 0
3 years ago
What is a way that i can get 3x and 2x to cancel out in this problem? i need to get - 5y = -13 and + 15y = -3 alone.
finlep [7]

Answer:

For 3x + 5y = -13 the answer is x= -13/3 + 5y/3

and for 2x + 15y=3 the answer x= 1.5 + (-15y/2)

Step-by-step:

For 3x+5y=-13 you have to add 5y on both sides. Then you have to divide by 3 on both sides and cancel out the common factor of 3. Then you would get x= 13/3 + 5y/3.

For 2x + 15y=3 you have to subtract 15y on both sides of the equation. You should have 2x=3 + (-15y). Then divide by 2 on both sides. 3 divided by 2 is 1.5 and -15y divided by 2 is -15y/2 since 2 doesn't have the same variable. so the answer should be x= 1.5 + (-15y/2).

6 0
3 years ago
A shareholders' group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least seven
Likurg_2 [28]

Complete question is;

A shareholders’ group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least seven years. A survey of companies reported in The Wall Street Journal found a sample mean tenure of x¯ = 7.27 years for CEOs with a standard deviation of s = 6.38 years (The Wall Street Journal, January 2, 2007).

a. Formulate hypotheses that can be used to challenge the validity of the claim made by

the shareholders’ group.

b. Assume 85 companies were included in the sample. What is the p-value for your hypothesis

test?

Answer:

A) Null hypothesis: H0: μ ≥ 7

Alternative hypothesis: μ < 7

B) p-value = 0.3488

Step-by-step explanation:

A) We are told that the shareholders group claimed that the mean tenure for a chief executive office (CEO) was at least seven years.

Thus, defining the hypothesis, we have;

Null hypothesis: H0: μ ≥ 7

Alternative hypothesis: μ < 7

B) Sample mean: x¯ = 7.27 years standard deviation: s = 6.38 years

Sample size = 85

Formula for the test static is;

t = (x¯ - μ)/(s/√n)

t = (7.27 - 7)/(6.38/√85)

t = 0.39

From online p-value from t score calculator attached using: t = 0.39, DF= 85 - 1 = 84, significance value = 0.05, one tail, we have;

p-value = 0.3488

5 0
3 years ago
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