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valkas [14]
3 years ago
10

Find the probability of spending a 5 and rolling a 1, 2 or 3.

Mathematics
1 answer:
Deffense [45]3 years ago
7 0
B because they all have an equal chance
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Last one on test Parallelogram ABCD is a rectangle.
marusya05 [52]
Answer: 10

2AX=BD
2(3y-5)=5y
6y-10=5y
add 10 to each side 6y+10-10=5y+10
6y=5y+10
subtract 5y from each side 6y-5y=5y-5y+10
6y-5y=y
y=10



3 0
3 years ago
Read 2 more answers
3cis(5pi/4) in polar form
madreJ [45]
Rectangular form:
z = -2.1213203-2.1213203i

Angle notation (phasor):
z = 3 ∠ -135°

Polar form:
z = 3 × (cos (-135°) + i sin (-135°))

Exponential form:
z = 3 × ei (-0.75) = 3 × ei (-3π/4)

Polar coordinates:
r = |z| = 3 ... magnitude (modulus, absolute value)
θ = arg z = -2.3561945 rad = -135° = -0.75π = -3π/4 rad ... angle (argument or phase)

Cartesian coordinates:
Cartesian form of imaginary number: z = -2.1213203-2.1213203i
Real part: x = Re z = -2.121
Imaginary part: y = Im z = -2.12132034
8 0
3 years ago
How many ways could you draw a single card from a 52 card deck and get either a heart or a king?
alekssr [168]
25 is the correct answer
3 0
3 years ago
Solve:<br>log2 3√(2)/ 16 = s​
jek_recluse [69]

Answer:

log1/8

Step-by-step explanation:

3 0
4 years ago
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​,and a standard deviation given by
Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with  

\mu = 62.5 in

\sigma = 2.2 in

the normal probability density function is given by  

f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}, then

(a) P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx = 0.5899

   (in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking P(\bar{X} < 63) where n = 37. \bar{X} is normally distributed with mean 62.5 in and standard deviation 2.2/\sqrt{37}. So, the probability density function is given by

g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}, and

P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0
3 years ago
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