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3 years ago

Replace the box with a number to make blank over 24 greater than 1 over 4 a true statement

Mathematics

2 answers:

ANTONII [103]3 years ago

6 0

Answer:
x>6
Hope I helped

PilotLPTM [1.2K]3 years ago

3 0

X/24>1/4
24 is 6 times 4, so 1 times 6=6
x>6 is the answer.

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3 years ago

HELP PLEASE!!

pogonyaev

Answer:

You need to have both the opposite and adjacent side of the right angled triangle. Tan = Opp/Adj

Step-by-step explanation:

Tangents are pretty much everywhere. A good example is maybe a slide at the playground. You’d need the tangent to figure out the angle of the slide and how it is.

The radius from the center of a circle to the point of tangency point shows that it would be perpendicular to the tangent line considering that anything with a radius is circular and the tangent line is… a line, making it impossible to be parallel.

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3 years ago

Travis has a budget of $300 that he can spend on perennial flowers and at least 6 annual flowers. Perennial flowers are 18 dolla

boyakko [2]

Answer:

18x + 15y ≤ 300
x ≥ 0; y ≥ 6

Step-by-step explanation:

Variables are defined in the problem statement.

18x +15y ≤ 300 . . . . total budget

y ≥ 6 . . . . . . . . . . . . minimum number of annuals

x ≥ 0 . . . . . number of perennials cannot be negative

This system of inequalities describes the situation.

6 0

3 years ago

he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02

Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

$P(k)= \frac{0.2^ke^{-0.2}}{k!}$

In this case, we are looking for P(0)

$P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%$

So, the probability that a car has no flaws is 98.01%

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and

$P(k)= \frac{2^ke^{-2}}{k!}$

and

$P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%$

And the probability that 10 cars have no flaws is 13.53%

Here, we are looking for P(1) with P defined as in b)

$P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%$

Hence, the probability that at most one car has no flaws is 27.06%

6 0

3 years ago

Read 2 more answers

3. A company with loud machinery needs to cut its sound intensity to 44% of its original level. By how many decibels would the l

sergeinik [125]

Answer: 3. The correct option is: A) 3.57 decibels.

6. The correct option is: B) $ln(\frac{x^2}{c^5})$ 

Step-by-step explanation:

3. Sound intensity is cut to 44% of its original level. That means, if the original sound intensity was 100, then now sound intensity will be 44.

So, $I_{0}= 100$ and $I= 44$

Using the formula $L= 10log(\frac{I}{I_{0}})$ , we will get.....

$L= 10log(\frac{44}{100})\\ \\ L= 10log(0.44)\\ \\ L= 10(-0.3565...)=-3.565...\approx -3.57$

(Rounded to the nearest hundredth)

So, the loudness would be reduced by 3.57 decibels.

6. Given expression is: $2ln(x) - 5ln(c)$

First applying the property $m*log(x)= log(x^m)$ , we will get.....

$ln(x^2)-ln(c^5)$

Now using the formula, $log(a)-log(b)= log(\frac{a}{b})$ , we will get....

$ln(x^2)-ln(c^5) = ln(\frac{x^2}{c^5})$

Thus, the answer as a single natural logarithm is $ln(\frac{x^2}{c^5})$

8 0

3 years ago