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Mrac [35]
2 years ago
7

A construction contractor received two deliveries of building supplies from a lumberyard. The two deliveries included 10 boxes o

f nails, which cost a total of $110.
Mathematics
1 answer:
hichkok12 [17]2 years ago
6 0
Missing question:
Determine n, the cost per box of nails.

Solution;
n = Total cost/Number of boxes
Total cost = $110
Total number of boxes in the two deliveries = 10 boxes.

Therefore,
n = $110/10 = $11
You might be interested in
Please help me with this question
natulia [17]

what's up

Seek understand, to be understood!

To get the answer we need the formula or source.

The question in another way if 4cm=1.2km, what is the 10cm=--------

first, you have to use the crisscross.

4cm=1.2km

10cm= x it is variable  

then, 4xx= 1.2kmx10cm

4x=12km

4x/4= 12/4

x=3km  


4 0
2 years ago
What is the volume of the pyramid?
s2008m [1.1K]

Answer:

16\sqrt{3} cm^{3}

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one. Please have a look at the attached photo,

<em>A solid oblique pyramid has an equilateral triangle as a base with an edge length of 4StartRoot 3 EndRoot cm and an area of 12StartRoot 3 EndRoot cm2. </em>

<em>What is the volume of the pyramid?</em>

My answer:

As we know, The volume of a pyramid = \frac{1}{3}base area × its height

Given:

  • Side lenght of the base is; 4\sqrt{3} cm

=> The area of the base is 12\sqrt{3} cm^{2}

  • In Δ ACB measure of angle ACB is 90° and m∠ CAB is 30°

We use: tan(30) = \frac{BC}{AC}

<=> BC = 4\sqrt{3}*tan(30)

= 4 cm

And BC is the height of the the pyramid

=> The volume of a pyramid = \frac{1}{3} 12\sqrt{3} cm^{2} * 4 cm

= 16\sqrt{3} cm^{3}

5 0
2 years ago
Read 2 more answers
Ramona's backyard is fenced and represented by parallelogram YARD with measures in meters. She installed a 12-meters fence to se
Alisiya [41]
The fencing line x is the height of a rectangle triangle of base = y, hypothenuse of 9 m, so we use Pythagoras theorem to solve:

hyp^2 = height^2 + base^2
9^2 = x^2 + y^2
x^2 = 81 - y^2


we can see that x is also the height of another rectangle triangle of base = 15 - y, hypothenuse of 12 m, so we use Pythagoras theorem to solve:
hyp^2 = height^2 + base^2
12^2 = x^2 + (15 - y)^2

lets expand:
144 = x^2 + 225 - 30y + y^2

substitute x^2 from the first equation in the last:
144 = 81 - y^2 + 225 - 30y + y^2
144 = 81 + 225 - 30y
30y = -144 + 81 + 225
y = 5.4 m

substitute in the fence equation:
x^2 = 81 - y^2
x^2 = 81 - 5.4^2
x = 7.2 m that is the length of the fence



3 0
2 years ago
The differential equation y′′=0 has one of the following two parameter families as its general solution: yyyy=C1ex+C2e−x=C1cos(x
svet-max [94.6K]

Answer:

y_c = 2 + 10*x

Step-by-step explanation:

Given:

                                                y'' = 0

Find:

- The solution to ODE such that y(0) = 2, y'(0) = 10

Solution:

- Assuming a solution y = Ce^(mt)

So,                                y' = C*me^(mt)

                                    y'' = C*m^2e^(mt)

- Back substitute into given ODE, we get:

                                    y'' = C*m^2e^(mt) = 0

                                    e^(mt) can not be equal to zero

- Hence,                       m^2 = 0

                                     m = 0 , 0 - (repeated roots)

- The complimentary function for repeated roots is:

                                    y_c = (C1 + C2*x)*e^(m*t)

                                    y_c = C1 + C2*x  

- Evaluate @ y(0) = 2

                                    2 = C1 + C2*0

                                    C1 = 2

-Evaluate @ y'(0) = 10

                                    y'(t) = C2 = 10

Hence,                         y_c = 2 + 10*x

5 0
2 years ago
(a) (i) Find the probability of getting at least one 3 when 9 fair dice are thrown. (ii) When n fair dice are thrown, the probab
Anettt [7]

Answer:

(a)

(i) The probability of getting at least one 3 when 9 fair dice are thrown is 0.8062.

(ii) The value of <em>n</em> is 12.

(b) The probability that Ronnie wins the game is 0.3572.

Step-by-step explanation:

(a)

(i)

The probability of getting a 3 on a single die roll is, p=\frac{1}{6}.

It is provided that <em>n</em> = 9 fair dice are thrown together.

The outcomes of each die is independent of the others.

The random variable <em>X</em> can be defined as the number of die with outcome as 3.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and p=\frac{1}{6}.

Compute the probability of getting at least one 3 as follows:

P(X\geq 1)=1-P(X=0)

                =1-[{9\choose 0}(\frac{1}{6})^{0}(1-\frac{1}{6})^{9-0}]\\\\=1-(\frac{5}{6})^{9}\\\\=1-0.19381\\\\=0.80619\\\\\approx 0.8062

Thus, the probability of getting at least one 3 when 9 fair dice are thrown is 0.8062.

(ii)

It is provided that:

P (X ≥ 1) > 0.90

Compute the value of <em>n</em> as follows:

P (X \geq 1) > 0.90\\\\1-(\frac{5}{6})^{n}>0.90\\\\(\frac{5}{6})^{n}

Thus, the value of <em>n</em> is 12.

(b)

It is provided that the bag contains 5 green balls and 3 yellow balls.

Ronnie and Julie play a game in which they take turns to draw a ball from the bag at random without replacement.

The winner of the game is the first person to draw a yellow ball.

Also provided that Julie draws the first ball.

P (Ronnie Wins) = P (The 1st yellow ball is selected at an even draw)

                          = P (The 1st yellow ball is drawn at 2nd, 4th and 6th draw)

                          = P (1st yellow ball is drawn at 2nd)

                                  + P (1st yellow ball is drawn at 4th)

                                        + P (1st yellow ball is drawn at 6th)

                          =[\frac{5}{8}\times \frac{3}{7}]+[\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}]+[\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}\times \frac{1}{4}\times 1]\\\\=0.2679+0.0714+0.0179\\\\=0.3572

Thus, the probability that Ronnie wins the game is 0.3572.

8 0
2 years ago
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