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3 years ago

Help help help help pls!

Mathematics

2 answers:

IgorLugansk [536]3 years ago

6 0

Option 2 is the correct answer! :)))

tensa zangetsu [6.8K]3 years ago

3 0

6/25 is your answer because you just add them up and take the 6

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How many vertices are in a polyhedron with 6 triangular faces?

Bingel [31]

I believe it will be 36 Im sorry if im wrong.

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3 years ago

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Which is the best estimate of each percent? 25% of 395

NemiM [27]

First, you should say 395*25 which is 9875. Then you divide:
9875/100= 98.75

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3 0

3 years ago

Solve. −35x+15>720 Drag and drop a number or symbol into each box to show the solution.

vampirchik [111]

Answer:

The solution of the inequality - 35x + 15 > 720 is $x$

Step-by-step explanation:

Consider the given inequality

- 35x + 15 > 720

Subtract both side by 15 , we get

- 35x + 15 - 15 > 720 - 15

- 35x > 705

Multiply both sides by -1 (Reverse the inequality )

(-35x)(-1) < 705 (-1)

35x < -705

Now, divide both side by 35

$\frac{35x}{35}=\frac{-705}{35}$

$\Rightarrow x=\frac{-141}{7}$

Thus, the solution of the inequality - 35x + 15 > 720 is $x$

6 0

3 years ago

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: $P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}$

Where $\lambda$ is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

$P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}$

Substitute the respective values in the above formula.

$P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})$

$P(k > 2)= 1 - (0.44596)$

$P(k > 2)=0.55404$

Hence, the required probability is 0.55404.

4 0

3 years ago

Round 5049 correct to 1 significant figure

Step2247 [10]

5000

Addition (+) and subtraction (-) round by the least number of decimals.
Multiplication (* or ×) and division (/ or ÷) round by the least number of significant figures.
Logarithm (log, ln) uses the input's number of significant figures as the result's number of decimals.
Antilogarithm (n^x.y) uses the power's number of decimals (mantissa) as the result's number of significant figures.
Exponentiation (n^x) only rounds by the significant figures in the base.
To count trailing zeros, add a decimal point at the end (e.g. 1000.) or use scientific notation (e.g. 1.000 × 10^3 or 1.000e3).
Zeros have all their digits counted as significant (e.g. 0 = 1, 0.00 = 3).
Rounds when required, after parentheses, and on the final step.

- BRAINLIEST answerer ❤️

7 0

2 years ago

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