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3 years ago

hey girlies im bored and need friends add my discord fruitloops#5035 anyone is welcomed. Also enjoy the free points

Mathematics

2 answers:

iren2701 [21]3 years ago

6 0

Answer:

Step-by-step explanation:

Virty [35]3 years ago

3 0

Answer:

Step-by-step explanation:

ok and thank you for the free points

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Aaron is standing at point C, watching his friends on a

igor_vitrenko [27]

Answer: 194

Step-by-step explanation:

Did the assignment

3 0

3 years ago

Read 2 more answers

1) Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 5 i

Luba_88 [7]

Answer:

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 70 and 72 inches tall? (Round your answer to four decimal places.)

z1 = (70-71)/4 = -0.25

z2 = (72-71/4 = 0.25

P(70<X<72) = p(-0.25<z<0.25) = 0.1974

Answer: 0.1974

(b) If a random sample of thirteen 18-year-old men is selected, what is the probability that the mean height x is between 70 and 72 inches? (Round your answer to four decimal places.)

z1 = (70-71)/(4/sqrt(13)) = -0.9014

z2 = (72-71/(4/sqrt(13)) = 0.9014

P(70<X<72) = p(-0.9014<z<0.9014) = 0.6326

Answer: 0.6326

please mark me the brainiest

3 0

2 years ago

A fair die is rolled 10 times. Find the expected value of:a) the sum of the numbers in the ten rolls;b) the number of multiples

velikii [3]

Answer:

a) 3.5

b) 3.33

c) $6 - 6\left(\begin{array}{ccc}5\\6\end{array}\right)^{10}$

Step-by-step explanation:

As given,

A fair die is rolled 10 times

Expected value of Sum of the number in 10 rolls = $\frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6}$

= 3.5

∴ we get

Expected value of Sum of the number in 10 rolls = 3.5

Ley Y : number of multiples of 3

Y be Binomial

Y - B(n = 10, p = $\frac{2}{6}$ )

Now,

Expected value = E(Y) = np = 10× $\frac{2}{6}$ = 3.33

Let m = total number of faces in a die

⇒m = 6

As die is roll 10 times

⇒n = 10

Now,

Let Y = number of different faces appears

Now,

Expected value, E(Y) = m - m $\left(\begin{array}{ccc}m-1\\m\end{array}\right)^{n}$

= $6 - 6\left(\begin{array}{ccc}6-1\\6\end{array}\right)^{10} = 6 - 6\left(\begin{array}{ccc}5\\6\end{array}\right)^{10}$

5 0

3 years ago

The equations are given below.

amid [387]

Answer:

neither

Step-by-step explanation:

they are parallel when they have the same slope and perpendicular when the slope is completely different

5 0

3 years ago

What is the mass of a solid ball of iron with a radius equal to 18 cm if it is known that the density of iron is 7800 kg/m

Shtirlitz [24]

Answer:

60.6528 kg

Step-by-step explanation:

Assuming the solid ball of iron is a sphere with a radius of 18 cm (or 0.18 m), we can compute the volume of the sphere:

$V=\frac{4}{3}\pi r^3\\\\V=\frac{4}{3}\pi (\frac{18}{100})^3\\\\V=\frac{4}{3}\pi (0.005832)\\\\V=0.007776$

Therefore, the volume of the solid ball is 0.007776 m³

Since we know that density is mass/volume, then we can find the mass of the iron:

$\rho=\frac{mass}{volume}\\ \\7800\: kg/m^3=\frac{mass}{0.007776\: m^3}\\\\mass=60.6528\:kg$

Thus, the mass of the solid ball of iron is 60.6528 kg

8 0

3 years ago