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3 years ago

PLEASE HELPPPPP

Mathematics

1 answer:

AnnyKZ [126]3 years ago

3 0

An even prime number would be 4,8,10

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You know that the average college student eats 0.75 pounds of food at lunch. If the standard deviation is 0.2 pounds of food, th

Fynjy0 [20]

Answer: $53.95

Step-by-step explanation:

Given: $\mu=0.75$ pounds

$\sigma= 0.2$ pounds

n= 50

Two tailed critical z-value for 90% confidence level = 1.645

Let x be the total amount of food that a cafeteria should have on hand to be 90 percent confident that it will not run out of food .

z-score = $\dfrac{x-\mu}{\sigma}$ $=\dfrac{x-0.75}{0.2}$

Then, $\dfrac{x-0.75}{0.2}=1.645$

$x-0.75=1.645\times0.2\\\\\Riightarrow\ x-0.75=0.329\\\\\Rightarrow\ x= 0.329+0.75\\\\\Rightarrow\ x= 1.079$

Foe 50 students, total amount of food = 1.079 x 50 = $53.95

hence, the cafeteria should have food of amount $53.95 so that it is be 90 percent confident that it will not run out of food.

7 0

4 years ago

The second term in a geometric sequence is 50. The fourth term in the same sequence is 112.5 . What is the common ration in this

ale4655 [162]

Let the common ratio be r.
then
a2*r*r=a4
rearrange,
r^2=a4/a2=112.5/50=2.25
r=sqrt(2.25)=1.5 = common ratio

7 0

3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

4 years ago

Helppp!!!! please!!!

Deffense [45]

Answer:

f finally

Step-by-step explanation:

hope it will helps you

6 0

3 years ago

Read 2 more answers

In answering the following questions, assume that the angular acceleration is constant and nonzero: alpha not equal to zero.

Ronch [10]

I think the correct answer would be true. The quantity represented by theta is a function of time. This is because the rate of change of angular speed is constant which means the angular distance is changing with time. Hope this answers the question. Have a nice day.

8 0

3 years ago