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3 years ago

5. Jill buys two circular pizzas - a small and a medium. The small pizza has

1 answer:

4 0

The area of small pizza is 2.778 times greater than the area of medium pizza, if the small pizza has a diameter of 6 inches and the medium pizza has a diameter of 10 inches.

Step-by-step explanation:

The given is,

The small pizza has a diameter of 6 inches.

The medium pizza has a diameter of 10 inches.

$\pi = 3.14$

Step: 1

Formula to calculate the area of circular pizza,

$A =\pi r^{2}$ ............................(1)

Where, r - Radius of pizza

Step: 2

Area of medium pizza,

Where, Diameter of medium pizza, D = 10 inches

Radius , $r = \frac{d}{2} = \frac{10}{2} = 5 inches$

r = 5 inches

Equation (1) becomes,

$A_{Medium} =\pi 5^{2}$

= (3.14)(25)

= 78.5

$A_{medium} = 78.5$ square inches

Step: 3

Area of small pizza,

Where, Diameter of small pizza, D = 6 inches

Radius , $r = \frac{d}{2} = \frac{6}{2} = 3 inches$

r = 3 inches

Equation (1) becomes,

$A_{Medium} =\pi 3^{2}$

= (3.14)(9)

= 28.26

$A_{medium} = 28.26$ square inches

Step: 4

Ratio of medium pizza to small pizza in square inches,

$= \frac{ A_{Medium } }{ A_{small} }$

$= \frac{78.5}{28.26}$

= 2.778

Area of medium pizza 2.778 times greater than the area of small pizza.

Result:

The area of small pizza is 2.778 times greater than the area of medium pizza.

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Hello:) anyone able to explain how to solve the equestrian for part (d) ? Thank youu

Anna007 [38]

Answer:

see explanation

Step-by-step explanation:

Given

4 $a^{4}$ - 5a² + 1 = 0

Use the substitution u = a², then equation is

4u² - 5u + 1 = 0

Consider the product of the coefficient of the u² term and the constant term

product = 4 × 1 = 4 and sum = - 5

The factors are - 4 and - 1

Use these factors to split the u- term

4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )

4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term

(u - 1)(4u - 1) = 0

Equate each factor to zero and solve for u

u - 1 = 0 ⇒ u = 1

4u - 1 = 0 ⇒ 4u = 1 ⇒ u = $\frac{1}{4}$

Convert u back into terms of a, that is

a² = 1 ⇒ a = ± 1

a² = $\frac{1}{4}$ ⇒ a = ± $\frac{1}{2}$

Solutions are a = ± 1 , a = ± $\frac{1}{2}$

7 0

3 years ago

Simplify the expression cos x cot x+ sin x please select the best answer from the choices provided a.0, b.csc x, c. Tan x, d sec

expeople1 [14]

Answer:

$cot x = \frac{cos x}{sin x}$

$cos x \frac{cos x}{sin x} + sin x$

$\frac{cos^2 x}{sin x} +sin x$

$sin^2 x + cos^2 x =1$

Solving for $cos^2 x$ we got $cos^2 x =1 -sin^2 x$ and replacing this we got:

$\frac{1-sin^2 x}{sin x} +sin x$

$\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x$

$csc x -sin x + sin x = csc x$

And then the best option for this case would be:

b.csc x

Step-by-step explanation:

For this case we have the following expression given:

$cos x cot x + sin x$

We know from math properties that the definition for cot is $cot x = \frac{cos x}{sin x}$

If we use this definition we got:

$cos x \frac{cos x}{sin x} + sin x$

$\frac{cos^2 x}{sin x} +sin x$

Now we can use the following identity:

$sin^2 x + cos^2 x =1$

Solving for $cos^2 x$ we got $cos^2 x =1 -sin^2 x$ and replacing this we got:

$\frac{1-sin^2 x}{sin x} +sin x$

$\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x$

$csc x -sin x + sin x = csc x$

And then the best option for this case would be:

b.csc x

4 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .