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2 years ago

13

On babylonian tablet ybc 4652, a problem is given that translates to this equation: x (x/7) (1/11) (x (x/7)) = 60 what is the so

lution to the equation? x = 48.125 x = 52.5 x = 60.125 x = 77

2 answers:

Yanka [14]2 years ago

5 0

Thanks for posting your question here. The answer to the above problem is x = <span>48.125. Below is the solution:
</span>
x+x/7+1/11(x+x/7)=60
x = x/1 = x • 7/7
x <span>• 7 + x/ 7 = 8x/7 - 60 = 0
</span>x + x/7 + 1/11 <span>• 8x/7 - 60 = 0
</span>8x <span>• 11 + 8x/ 77 = 96x/ 77
</span>96x - 4620 = 12 <span>• (8x-385)
</span>8x - 385 = 0
x = 48.125

mart [117]2 years ago

3 0

Answer:

Option (a) is correct.

x = 48.125

Step-by-step explanation:

Given: $x+\frac{x}{7}+ \frac{1}{11}(x+\frac{x}{7})=60$

We have to solve for x,

Consider the given expression $x+\frac{x}{7}+ \frac{1}{11}(x+\frac{x}{7})=60$

First solving for brackets, we get,

$x+\frac{x}{7}=\frac{7x+x}{7}=\frac{8x}{7}$

Put , we get,

$x+\frac{x}{7}+ \frac{1}{11}(\frac{8x}{7})=60$

Simplify, we have,

$x+\frac{x}{7}+(\frac{8x}{77})=60$

Taking LCM(7,77) = 77

We have,

$\frac{77x+11x+8x}{77}=60$

Simplify, we have,

$\frac{96x}{77}=60$

Multiply both side by 77, we have,

$96x = 60 \times 77$

$96x =4620$

Divide both side by 96, we have,

$x=\frac{4620}{96}=48.125$

Thus, x = 48.125

Option (a) is correct.

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A water trough has two congruent isosceles trapezoids as ends and two congruent rectangles as sides.

BlackZzzverrR [31]

Answer:

Part a) The exterior surface area is equal to $160\ ft^{2}$

Part b) The volume is equal to $240\ ft^{3}$

Part c) The volume water left in the trough will be $84\ ft^{3}$

Step-by-step explanation:

Part a) we know that

The exterior surface area is equal to the area of both trapezoids plus the area of both rectangles

so

<em>Find the area of two rectangles</em>

$A=2[12*5]=120\ ft^{2}$

<em>Find the area of two trapezoids</em>

$A=2[\frac{1}{2}(8+2)h]$

Applying Pythagoras theorem calculate the height h

$h^{2}=5^{2}-3^{2}$

$h^{2}=16$

$h=4\ ft$

substitute the value of h to find the area

$A=2[\frac{1}{2}(8+2)(4)]=40\ ft^{2}$

The exterior surface area is equal to

$120\ ft^{2}+40\ ft^{2}=160\ ft^{2}$

Part b) Find the volume

We know that

The volume is equal to

$V=BL$

where

B is the area of the trapezoidal face

L is the length of the trough

we have

$B=20\ ft^{2}$

$L=12\ ft$

substitute

$V=20(12)=240\ ft^{3}$

Part c)

<em>step 1</em>

Calculate the area of the trapezoid for h=2 ft (the half)

the length of the midsegment of the trapezoid is (8+2)/2=5 ft

$A=\frac{1}{2}(5+2)(2)=7\ ft^{2}$

<em>step 2</em>

Find the volume

The volume is equal to

$V=BL$

where

B is the area of the trapezoidal face

L is the length of the trough

we have

$B=7\ ft^{2}$

$L=12\ ft$

substitute

$V=7(12)=84\ ft^{3}$

4 0

2 years ago

4(g - 1 )=24<br>what is the answer

babunello [35]

Hello There!

First Expand:
4g - 4 = 24
Then solve:
4g = 24 + 4
4g = 28
g = 28/4
g = 7.

Hope This Helps You!
Good Luck :)

6 0

2 years ago

Read 2 more answers

One way to measure whether the trees in the Wade Tract are uniformly distributed is to examine the average location in the north

tatyana61 [14]

Answer:

Null hypothesis be H₀ : μ = 100

Alternative hypothesis Hₐ : μ < 100

There is sufficient statistical evidence to suggest that the average location of Wade Tract is 100

Step-by-step explanation:

Here we have;

Let our null hypothesis be H₀ : μ = 100 Average location of trees in the Wade Tract is 100

Our alternative hypothesis becomes Hₐ : μ < 100 at 95% confidence level

Proposed average location in Wade Tract, μ = 100

Sample mean, $\bar x$ = 99.74

Standard deviation, s = 58

Sample size, n = 584

The t test formula is therefore;

$t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}$

Therefore, with df = 584 -1 = 583, and α = (1 - 0.95)/2 = 0.025

We have $t_{\alpha /2}$ = -1.65

Plugging in the values into the t test formula, we have t = -0.108338, from which the p-value is given as p = 0.4569 which is much more than the value of α, therefore, we accept the null hypothesis as follows;

There is sufficient statistical evidence to suggest that the average location of Wade Tract = 100.

7 0

2 years ago

The steps below show the work of a student used to calculate the number of yards in 804.5 meters.

Butoxors [25]

Answer:

see below

Step-by-step explanation:

804.5 meters. to yards

Step 1:

(1 mile = 1,609 meters)

(1 mile = 1,760 yards)

1,609 meters is being multiplied by conversion factor 1 mile over 804.5 meters equals 2 miles

<u>1,609 meter </u> x <u> 1 mile </u> = 2 miles

804.5 meters

Step 2:

(1 mile = 1,609 meters)

(1 mile = 1,760 yards)

2 miles is being multiplied by conversion factor 1760 yards over 1 mile equals 3,520 yards

2 miles x <u> 1,760 yards </u> = 3,520 yards

1 mile

no need of step 3. the conversion below should be on step 2.

to convert miles into yards.

(1 mile = 1,609 meters)

(1 mile = 1,760 yards)

3 0

2 years ago

let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo

anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

$f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\$

a) P(x > 5) = $\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865$

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = $\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}$

But P(x > a) = 0.05

Therefore:

$e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98$

a ≅ 12

6 0

2 years ago