A classmate is starting a fitness program

DiKsa [7]

2400 because it is only 93 away from 2493 and 2353 is more than 93 away

8 0

3 years ago

Use the Central Limit Theorem to find a mean given a probability Question A video game company sells an average of 132 games a m

aev [14]

Answer:

The number of games must a store sell in order to be eligible for a reward is 135.

Step-by-step explanation:

Let the random variable X represent the number of video games sold in a month by the sores.

The random variable X has a mean of, μ = 132 and a standard deviation of, σ = 9.

It is provided that the company is looking to reward stores that are selling in the top 7%.

That is, $P (\bar X > \bar x) = 0.07$ .

The z-score related to this probability is, z = 1.48.

Compute the number of games must a store sell in order to be eligible for a reward as follows:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}$

$\bar x=\mu+z\cdot \sigma/\sqrt{n}$

$=132+1.48\times (9/\sqrt{36})\\\\=132+2.22\\\\=134.22\\\\\approx 135$

Thus, the number of games must a store sell in order to be eligible for a reward is 135.

5 0

3 years ago

Solve for z, wz-4/ 5z

saw5 [17]

$x=\dfrac{wz-4}{5z}\\\\5xz=wz-4\\\\5xz-wz=-4\\\\z(5x-w)=-4\\\\z=-\dfrac{4}{5x-w}$

7 0

4 years ago

I know that the surface area is equal to 264 but I need to know how to get there (the steps)

Ray Of Light [21]

You posted the question twice ... I have repeated the answer here .. .

Find the area of all the sides:
10 x 9 = 90
9 x 8 = 72
1/2 x 6 x 8 = 24
1/2 x 6 x 8 = 24
6 x 9 = 54

Add them up:
90 + 72 + 24 + 24 + 54 = 264 square metres

7 0

3 years ago

Read 2 more answers

Given: Tangents LA and LB, m∠AOB=110°. Find: m∠ALO.

Yuri [45]

Answer:

The measure of angle ALO = 35°

Step-by-step explanation:

* Lets talk about the tangents to a circle drawn from a point

outside the circle:

- They are equal in length

- They are perpendicular to the radii at the point of tang-ency

- If we join the points of tang-ency withe the center of the circle

and join the outside point with center of the circle, we

formed two congruent triangles

* Now lets check our problem

- AL and LB are two tangents to the circle O ant points A and B

∴ LA = LB ⇒ (1)

- OA and OB are radii in circle O

∴ OA = OB ⇒ (2)

∴ OA ⊥ AL and OB ⊥ BL ⇒ radius and tangent

∴ m∠OAL = 90° , m∠OBL = 90

∴ m∠OAL = m∠OBL ⇒ (3)

* From (1) , (2) , (3)

- The two triangles LAO and LBO are congruent ⇒ SAS

- SAS is one case of congruence

∵ The lengths of two sides and the measure of the including

angle between them in one triangle equal to the

corresponding sides and angles in the second triangle,

then the two triangles are congruent

∴ m∠LOA = m∠LOB

∵ m∠AOB = 110°

∴ m∠LOA = m∠LOB = 110/2 = 55°

* In ΔLAO

∵ m∠ LAO = 90° , m∠LOA = 55°

∴ m∠ALO = 180 - (90 + 55) = 180 - 145 = 35°

* The measure of angle ALO = 35°

8 0

4 years ago