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3 years ago

Can some one please help me with this problem?

Mathematics

2 answers:

Nostrana [21]3 years ago

8 0

(6×90)+(2×50)=$640

$640 was collected on opening night

Lady bird [3.3K]3 years ago

6 0

6a+2s=x (total amount of money)
6 (90)+2 (50)= x
540+100=640
The total amount of money is $640
hope that helps!

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PLEASE HELP ASAP !!

wariber [46]

Answer:

First figure: $954cm^3$

Second figure: $1,508yd^3$

Third figure:

Height= q
Side length = r

Fourth figure:  $726cm^3$

Explanation:

A. First figure:

1. Formula:

$\text{Volume of a cylinder}=\pi \times radius^2\times length$

2. Data:

radius = 9cm / 2 = 4.5cm

length = 15 cm

3. Substitute in the formula and compute:

$Volume=\pi \times (4.5cm)^2\times (15cm)\approx 954cm^3\approx 954cm^3$

B. Second figure

1. Formula:

$\text{Volume of a leaned cylinder}=\pi \times radius^2\times height$

2. Data:

radius = 12yd

height = 40 yd

3. Substitute and compute:

$Volume=\pi \times (12yd)^2\times (40yd)\approx 1,507.96yd^3\approx 1,508yd^3$

C) Third figure

a) The height is the segment that goes vertically upward from the center of the base to the apex of the pyramid, i.e. q .

The apex is the point where the three leaned edges intersect each other.

b) The side length is the measure of the edge of the base, i.e. r .

When the base of the pyramid is a square the four edges of the base have the same side length.

D) Fourth figure

1. Formula

The volume of a square pyramide is one third the product of the area of the base (B) and the height H).

$Volume=(1/3)B\times H$

2. Data:

height: H = 18cm

side length of the base: 11 cm

3. Calculations

a) Calculate the area of the base.

The base is a square of side length equal to 11 cm:

$\text{Area of the base}=B=(11cm)^2=121cm^2$

b) Volume of the pyramid:

$Volume=(1/3)B\times H=(1/3)\times 121cm^2\times 18cm=726cm^3$

4 0

3 years ago

Read 2 more answers

The thicknesses of 81 randomly selected aluminum sheets were found to have a variance of 3.23. Construct the 98% confidence inte

Yuliya22 [10]

Answer:

The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.

Step-by-step explanation:

The confidence interval for population variance is given as below:

$[(n - 1)\times S^{2} / X^{2} \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2} / X^{2} 1- \alpha/2, n- 1 ]$

We are given

Confidence level = 98%

Sample size = n = 81

Degrees of freedom = n – 1 = 80

Sample Variance = S^2 = 3.23

$X^{2}_{[\alpha/2, n - 1]} = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401$

(By using chi-square table)

[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]

[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]

2.3004 < σ^2 < 4.8263

Lower limit = 2.30

Upper limit = 4.83.

8 0

2 years ago

Eli sells refrigerators. His base salary is $250 per week, and his commission is $50 for each refrigerator (r) that he sells. He

Mekhanik [1.2K]

Answer:

C and e

Step-by-step explanation:

the 50r represents the commision he receives for each refrigerator he sells, and the 250 is the base salary and it’s bigger than, >, because he needs to earn MORE than 900 and you get D by solving the inequality, 50r >900, r>15

would be amazing if you marked this brainliest :)

5 0

3 years ago

Need help please, thanks!

MakcuM [25]

Given:
1 inch : 18 kilometers

Let us use the ratio and proportion method:

a:b = c:d where ad = bc

0.8 inches is 14.4 kilometers

1 : 18 = 0.8 : x
x = 0.8 * 18
x = 14.4 km

1.4 inches is 25.2 kilometers

1: 18 = 1.4 : x
x = 18 * 1.4
x = 25.2 km

2.1 inches is 37.8 kilometers

1 : 18 = 2.1 : x
x = 18 * 2.1
x = 37.8 km

4 0

3 years ago

Given that QRSTU=QXWVU, find

Mrac [35]

Angle UQR = Angle UQX since they are corresponding angles. Note how the top polygon reflects over the line QU to get the bottom polygon.

So because of this, and because angle UQX = 45, this means the answer is 45 degrees

7 0

3 years ago

Read 2 more answers