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Tamiku [17]
3 years ago
6

The garden has an area of 28 square meters. The length of the garden is 7 meters. What is the width of the garden?

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
8 0

Answer:

4 meters

Step-by-step explanation:

Area is length x width

28=7n where n is the unknown width

Solve for n and u get 4

GREYUIT [131]3 years ago
4 0

Answer:

4 metres

Step-by-step explanation:

The garden is in the form of a rectangle. The area of a rectangle is denoted by A = lw, where l is the length and w is the width.

Here, we're given the length is l = 7 and the area is A = 28, so we can plug these into the formula:

A = lw

28 = 7 * w

Divide both sides by 7:

w = 28 / 7 = 4

Thus, the width is 4 metres.

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2 years ago
Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra
Nana76 [90]

The length of BC is 4 \sqrt{5}.

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

\Rightarrow  16\times 5 = x\times x

\Rightarrow  80= x^2

\Rightarrow  16\times 5= x^2

Taking square root on both sides.

\Rightarrow  \sqrt{16\times 5} = \sqrt{x^2}

\Rightarrow  \sqrt{4^2\times 5} = \sqrt{x^2}

square and square roots get canceled, we get

\Rightarrow  4\sqrt{ 5} = x

The length of BC is 4 \sqrt{5}.

7 0
3 years ago
Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are
Musya8 [376]

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

7 0
3 years ago
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