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3 years ago

H+3h+4h=100 how do i get h?

Mathematics

2 answers:

Dominik [7]3 years ago

7 0

Answer:

h = 12.5

Step-by-step explanation:

Simplifying

h + 3h + 4h = 100

Combine like terms: h + 3h = 4h

4h + 4h = 100

Combine like terms: 4h + 4h = 8h

8h = 100

Solving

8h = 100

Solving for variable 'h'.

Move all terms containing h to the left, all other terms to the right.

Divide each side by '8'.

h = 12.5

Simplifying

h = 12.5

rewona [7]3 years ago

6 0

Answer:

h=12.5

Step-by-step explanation:

To solve for h, we need to isolate the variable

h+3h+4h =100

Combine like terms by adding h, 3h and 4h

8h=100

Divide both sides by 8 to "reverse" the multiplication being done to h

8h/8=100/8

The 8s on the left will cancel, leaving us with:

h=12.5

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At a certain coffee shop, all the customers buy a cup of coffee and some also buy a doughnut. The shop owner believes that the n

wolverine [178]

Answer:

(1) The probability that the shop owner sells over 2000 cups of coffee in a week is 0.2514.

(2) The shop owner has no reasonable chance to expect earning a profit more than $300.

(3) The probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.

Step-by-step explanation:

Let X = number of cups of coffee sold and Y = number of donuts sold.

The random variable X follows a Normal distribution with parameters μ = 320 and σ = 20.

The random variable Y follows a Normal distribution with parameters μ = 150 and σ = 12.

The shop owner opens the shop 6 days a week.

(1)

Compute the probability that the shop owner sells over 2000 cups of coffee in a week as follows:

$P(X>2000)=P(\frac{X-\mu}{\sigma}>\frac{2000-(6\times320)}{6\times20})\\=P(Z>0.67)\\=1-P(Z$

Thus, the probability that the shop owner sells over 2000 cups of coffee in a week is 0.2514.

(2)

The equation representing the profit earned on selling 1 cup of coffee and 1 doughnut in a day is:

P = 0.5X + 0.4Y

Compute the probability that the shop owner earns more than $300 as profit as follows:

$P(Profit>300)=P(\frac{Profit-\mu}{\sigma}>\frac{300-((0.5\times320)+(0.4\times150))}{\sqrt{0.5^{2}(20)^{2}+0.4^{2}(12)^{2}}})\\=P(Z>7.21)\\\approx0$

The probability of earning a profit more then $300 is approximately 0.

Thus, the shop owner has no reasonable chance to expect earning a profit more than $300.

(3)

The expression representing the statement "he'll sell a doughnut to more than half of his coffee customers" is:

Y > 0.5X

Y - 0.5X > 0

Compute the probability of the event (Y - 0.5X > 0) as follows:

$P(Y - 0.5X > 0)=P(\frac{(Y - 0.5X) -\mu}{\sigma}>\frac{0-(150-(0.5\times320}{\sqrt{12^{2}+0.5^{2}20^{2}}})\\=P(Z>0.64)\\=1-P(Z$

Thus, the probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.

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3 years ago

A pair of tow trucks are pulling on a trailer that is stuck

Varvara68 [4.7K]

Answer:

R=10,244 Lb

Step-by-step explanation:

In order to solve this problem, we can start by drawing the situation. (See attached picture).

We can look at the forces applied by the trucks as vectors, so in this case, in order to find the resultant force, we can add the two vectors. You can do so by drawing one force after the other and join the start of the first force with the end of the second force to form a triangle. The missing side will be the resultant force. In that triangle I drew there, the inner angle will be found by subtracting 180°-110°=70°

In this case we can find the resultant force by using the law of cosines:

$A^{2}=B^{2}+C^{2}-2AB cos \alpha$

so we can use the data given by the problem:

$R^{2}=T_{1}^{2}+T_{2}^{2}-2T_{1}T_{2} cos \alpha$

so we can solve this for R, so we get:

$R=\sqrt{T_{1}^{2}+T_{2}^{2}-2T_{1}T_{2} cos \alpha}$

now we can substitute:

$R=\sqrt{(10,000Lb)^{2}+(7,500Lb)^{2}-2(10,000Lb)(7,500Lb) cos (70^{0})}$

which yields:

R=10,244Lb

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