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2 years ago

4. Is it possible to have a triangle with the sides 6.1 cm, 3.2cm, 2.4cm? Justify your reason with suitable reason.

1 answer:

3 0

Answer: No
Explanation: There is a rule that stays that a triangle's 2 least sides must add up to be greater than the biggest side. 3.2 plus 2.4 is not greater than 6.1.

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Verify the identity. Show your work. (1 + tan^2u)(1 - sin^2u) = 1

yuradex [85]

$\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\ 1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\\
[1+tan^2(u)][1-sin^2(u)]=1
\\\\\\\
[sec^2(u)][cos^2(u)]\implies \cfrac{1}{cos^2(u)}\cdot cos^2(u)\implies \cfrac{cos^2(u)}{cos^2(u)}\implies 1$

7 0

3 years ago

The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m

UkoKoshka [18]

Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of $\bar X$ Round your answers to two decimal places.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

4 0

3 years ago

Ill give brainliest - please help ASAP<br><br> WITH THE WORK THO PLEASE?

ohaa [14]

Answer:

we have.

11x=½(16x+16+50)

22x=16x+66

22x-16x=66

6x=66

x=66/6

x=11

118°=½(6x-11+181)

236=6x+170

6x=236-170

x=66/6

x=11°

34°=½(arcAK-arc LN)

34×2=(110-arc LN)

arc LN =110-68

arc LN=42°

<L=½(arc EN-arc KM)

<L=½(139°-73°)

<L=33

4 0

3 years ago

The graph of a proportional relationship contains the point (-30, 18)

Elena L [17]

Answer:

$k=-\frac{3}{5}$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

we have the point (-30,18)

x=-30, y=18

Find the value of k

$k=\frac{y}{x}$

substitute

$k=\frac{18}{-30}$

Simplify

Divide by 6 both numerator and denominator

$k=-\frac{3}{5}$

4 0

3 years ago

Select the correct answer.

icang [17]

Answer: C (0,2)

Step-by-step explanation: Y intercept is when the x value is equal to 0. Find the value in the table where the x value is 0 and then look at the h(x) value or the y value to find the answer. In this case the point where x=0 h(x) = 2.

7 0

3 years ago