Can someone help me plz need this done by tmrw

7b<br> Evaluate when b= -1<br> b-3

tiny-mole [99]

Answer:

b-3

-1-3

-4

is the correct answer

5 0

3 years ago

Find the fifth roots of 243(cos 240° + i sin 240°).

Mazyrski [523]

Answer:

See below.

Step-by-step explanation:

Fifth root of 243 = 3,

Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),

then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).

Equating equal parts and using de Moivre's theorem:

r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240

r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p

So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360

So there are 5 distinct solutions given by:

3(cos 48 + i sin 48),

3(cos 120 + i sin 120),

3(cos 192 + i sin 192),

3(cos 264 + i sin 264),

3(cos 336 + i sin 336).. (Answer).

7 0

3 years ago

A water balloon is tossed into the air with an upward velocity of 25 ft/s. Its height h(t) in ft after t seconds is given by the

sammy [17]

19. The balloon hits the ground means -16t2+25t+3=0
After you solve the equation, there are 2 solution ~1.67 or ~-0.11 (the time can not be negative) so the solution is 1.67s
20.substitute t=1 to the function -16(1)2+25(1)+3=-16+25+3=12 ft. So the height is 12ft when t=1

7 0

3 years ago

Read 2 more answers

The weights of certain machine components are normally distributed with a mean of 5.19 ounces and a standard deviation of 0.05 o

Brums [2.3K]

Answer:

The weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.

Step-by-step explanation:

We are given that

Mean, $\mu=5.19$

Standard deviation, $\sigma=0.05$

We have to find the two weights that separate the top 8% and the bottom 8%.

Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.

Z-value for p-value =0.08 = $-1.41$

For 8% bottom

$Z=\frac{x_1-\mu}{\sigma}=-1.41$

$\frac{x_1-5.19}{0.05}=-1.41$

$x_1-5.19=-1.41\times 0.05$

$x_1=-1.41\times 0.05+5.19$

$x_1=5.1195$

For 8% top

p-Value=1-0.08=0.92

Z- value=1.41

Now,

$\frac{x_2-5.19}{0.05}=1.41$

$x_2-5.19=1.41\times 0.05$

$x_2=1.41\times 0.05+5.19$

$x_2=5.2605$

(x1,x2)=(5.1195,5.2605)

4 0

3 years ago

Consider the probability that at least 88 out of 158 software users will not call technical support. Assume the probability that

Nookie1986 [14]

Answer:

$E(X) = 158*0.57 =90.06$

And the standard deviation for the random variable is given by:

$\sigma= \sqrt{158*0.57*(1-0.57)} = 6.223$

And the distribution for the approximation is given by:

$X \sim N (\mu = 90.06, \sigma = 6.223)$

We can use the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

And using the complement rule we got:

$P(X \geq 88) = 1-P(X$

And using the standard normal table we got:

$P(X \geq 88) = 1-P(X$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n = 158, p =0.57)$