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Hatshy [7]
2 years ago
6

The probability that it will snow tomorrow is 15%. What are the odds in favour of it snowing tomorrow

Mathematics
1 answer:
Elenna [48]2 years ago
6 0

Answer:

The probabilities of all possible outcomes must sum to 1, since something has to happen. So

Psnow=13⟹Pno snow=1−13=23.Psnow=13⟹Pno snow=1−13=23.

You might be interested in
Suppose the diameter of a circle is in inc what is the radius
Vikentia [17]

Answer:

pi4

Step-by-step explanation:

Formula is piD, diamter is 4, so pi4 which equals 12.56

BRAINYLIST

5 0
2 years ago
When Ximena commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 38 minutes and a
irga5000 [103]

Answer:

The interval that represents the middle 68% of her commute times is between 33.5 and 42.5 minutes.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 38 minutes, standard deviation of 4.5 minutes.

Determine the interval that represents the middle 68% of her commute times.

Within 1 standard deviation of the mean. So

38 - 4.5 = 33.5 minutes

38 + 4.5 = 42.5 minutes.

The interval that represents the middle 68% of her commute times is between 33.5 and 42.5 minutes.

6 0
2 years ago
14x + 64 = <br><br> x = 3<br><br> What is the answer
Nostrana [21]

x = 3.

Plug in 3 for x in the equation

14(3) + 64

Remember to follow PEMDAS. First, multiply 3 with 14

3 x 14 = 42

Finally, add 64

42 + 64 = 106

106 is your answer

hope this helps

5 0
3 years ago
Directions: Calculate the area of a circle using 3.14x the radius
Leokris [45]

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 1 \div 2

\qquad \sf  \dashrightarrow \:r = 0.5 \:  \: yd

Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

8 0
2 years ago
A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.
tatyana61 [14]

Answer:

(a)g(n)=0.05\cdot 2^n

(b)g^{-1}(n)=\log_{2}20g(n)

(c)43 times

Step-by-step explanation:

<u>Part A</u>

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n

<u>Part B</u>

<u />g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)<u />

<u />

<u>Part C</u>

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43

You must fold the paper 43 times to make the folded paper have a thickness that is the same as the distance from the earth to the moon.

3 0
2 years ago
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