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3 years ago

A coin is tossed 20 times. The outcomes H (for heads) and T (for tails) are recorded:

Mathematics

2 answers:

Viefleur [7K]3 years ago

6 0

Answer:

0.55

Step-by-step explanation:

lina2011 [118]3 years ago

4 0

After all 20 tosses of the coin, a tail was recorded 11 times. We can write this as 11/20.

To convert this into a decimal, we simply multiply the numerator and denominator by 5 so that the fraction is out of 100, giving us 55/100.
When we do the sum of 55 ÷ 100, we get the answer of 0.55.

I hope this helps!

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Use the graph to solve the system of linear equations. x+y=5 y – 2x = -4

Romashka [77]

Y=-x+5
Y=2x-4
X would be your slope
Y intercepts would be (0, 5) , ( 0,-4)
Formula i used was Y=Mx+B

3 0

2 years ago

A community swimming pool is a rectangular prism that is 30 feet long, 12 feet wide, and 5 feet deep. The wading pool is half as

topjm [15]

Answer:

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

Step-by-step explanation:

By definition of rectangular prism, we get the respective formulas for the volumes of the community swimming pool and the wadling pool, respectively:

Community swimming pool

$V_{c} = l\cdot w\cdot h (1)$

Wading pool

$V_{w} = \left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w (2)$

Where:

l – Length of the swimming pool, measured in feet.

h – Depth of the swimming pool, measured in feet.

w – Width of the swimming pool, measured in feet.

$V_{c}$ – Volume of the community swimming pool, measured in cubic feet.

$V_{w}$ – Volume of the wading swimming pool, measured in cubic feet.

The ratio of the volume of the community swimming pool to the volume of the wadling pool is:

$\frac{V_{c}}{V_{w}} = \frac{l\cdot w \cdot h}{\left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w} (3)$

$\frac{V_{c}}{V_{w}} = \frac{1}{\frac{1}{4} }$

$\frac{V_{c}}{V_{w}} = 4$

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

8 0

1 year ago

Read 2 more answers

Hello can some one please help me

Artemon [7]

It’s 4/25 because 2/5 times 2/5 the numerator is 2 times 2 which is four and the denominator is 5 times 5 which is 25

8 0

1 year ago

Read 2 more answers

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann

mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

The probability mass function of a Geometric distribution is:

$P(Y=y)=(1-p)^{y}p;\ y=0,12...$

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

$P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844$

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

$=(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780$

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable Y as follows:

$\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445$

Compute the standard deviation of the random variable Y as follows:

$\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88$

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

= P (Y ≥ 3.325)

= P (Y ≥ 3)

= 1 - P (Y < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064$

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0

2 years ago

Subtract (4x2 - x + 6) from (3x2 + 5x - 8).

bearhunter [10]

Step-by-step explanation:

$(4x^2-x+6)-(3x^2+5x-8)=4x^2-x+6-3x^2-5x+8$

By simplifying the right side of the equation, we come up with

$x^2+6x-14$ , or D

5 0

2 years ago