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4vir4ik [10]
3 years ago
7

PLZZZZZ HELP Two standard number cubes are thrown, one after the other. What is the probability that the first cube lands showin

g a 1 and the second cube lands showing a number that is not 1? Round the answer to the nearest thousandth.
A.
0.056
B.
0.139
C.
0.167
D.
0.278
Mathematics
2 answers:
Akimi4 [234]3 years ago
5 0

Answer:

Answer is b on edge202

Step-by-step explanation:

0.139

katen-ka-za [31]3 years ago
3 0

|\Omega|=6^2=36\\ |A|=1\cdot5=5\\\\ P(A)=\dfrac{5}{36}\approx0.139

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Please help me Find a, b, and c
zlopas [31]

Answer:

It is A.

Step-by-step explanation:

To solve for b, use the 45-45-90 triangle theorem, in which each of the legs is x, so the legs would be 8. The hypotenuse would therefore be 8√2.

So without further solving the answer is A, since it's the only one with 8√2.

However, I will still solve for A and C. Using the 30-60-90 theorem, we have the sides as x, x√3, and 2x. The second longest side is b. Using this, we find a = 4√6 and c to be 4√2

3 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
(4x1,000)+(5x100)+(9x10)+(2x1)+(7x1/10) in standard form
Igoryamba

Answer:

4592.7

Step-by-step explanation:

Add them up:

4000 + 500 + 90 + 2 + 0.7 = 4592.7

6 0
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Artyom0805 [142]
The square footage of the room is length times width...in this case 16 x 12 =192 then divide that by 7 and it can accommodate 27 people.
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Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Hint: the
Julli [10]

Answer:

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 16)n = 182, x = 135; 95 percent

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6 0
3 years ago
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