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3 years ago

Another one i need help with

Mathematics

1 answer:

Ugo [173]3 years ago

4 0

The length of each shape will be an even number.

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Let sintheta =2times square root 3/5 and pi/2

Gre4nikov [31]

Given:

$sin\theta=\frac{2\sqrt{3}}{5},\text{ and }\frac{\pi}{2}$

It's required to find cos 2θ.

We use the formula:

$cos(2\theta)=1-2sin^2\theta$

Substituting:

$cos(2\theta)=1-2(\frac{2\sqrt{3}}{5})^2$

Operating:

$\begin{gathered} cos(2\theta)=1-2*\frac{4*3}{25} \\ cos(2\theta)=1-\frac{24}{25} \\ cos(2\theta)=\frac{1}{25} \end{gathered}$

Answer: 1/25

3 0

1 year ago

The owner of a grocery store wants to mix two kinds of candy together to make 15 lb that he can sell for $5.00 per lb. He wants

jeyben [28]

Answer:

9lb chocolate candies

6lb sugar candies

Step-by-step explanation:

Yes, I also happen to go to RSM. (6b)

Equation: 7x+2(15-x)=5*15

Solve:

7x+(2)(15)+(2)(−x)=(5)(15)

7x+30+(−2x)=75

(7x+(−2x))+(30)=75

5x+30=75

5x=45

x=9lb chocolate candies

Now, calculate the sugar candies.

15-9=6lb sugar candies

6 0

3 years ago

Read 2 more answers

A apple pie is 2 oz. for 9.99 what is the unit rate

lora16 [44]

1 oz is 5 dollars rounded.

8 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago

Find the smallest possible integer m such that 60m is a square number.

noname [10]

Answer:

Step-by-step explanation:

8 0

2 years ago