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3 years ago

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

Mathematics

1 answer:

Elenna [48]3 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 ]

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<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>

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

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

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

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

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

Step-by-step explanation:

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

<u />

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