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tangare [24]
2 years ago
11

Amir is laying stone for his new patio. The diagram of the patio is shown. How many

Mathematics
1 answer:
lutik1710 [3]2 years ago
7 0

Answer: 108ft^2

Step-by-step explanation:

First, at the top we have two triangle rectangles.

The area of a triangle rectangle is equal to the product of their cathetus divided by two.

We know that one common cathetus for both triangles is 6ft.

And both share the bottom line that is 14ft long, so half of that corresponds to each triangle.

14ft/2 = 7ft

Then each triangle has a cathetus of 6ft and one of 7ft.

Then the area of each triangle rectangle at the top is:

6ft*7ft/2 = 21ft^2

And we have two of them, so the area for now is:

A1 = 21ft^2 + 21ft^2 = 42ft^2.

Now let's look at the bottom part, we can divide this into two triangle rectangles and one rectangle.

First, bottom side is 8ft, then the difference between this side and the 14ft one is:

14ft - 8ft = 6ft

(We calculate this because we are making a rectangle, then the bottom side length must be equal than the top one).

Then we have a surpass of 6ft, which we will divide into both triangle rectangles (3ft for each, this is one of the cathetus).

And we can see that the cathetus shown of this triangle rectangle is 6ft (this is also the other side of our rectangle)

Then we have two triangles with cathetus 6ft and 3 ft, the area is:

A = 6ft*3ft/2 = 9ft^2

And we have two of them, then A2 = 18ft^2.

And the rectangle is 8ft by 6ft, the area is:

A3 = 8ft*6ft = 48ft^2

Then if we add all the areas that we found, we have:

A1 + A2 + A3 = 42ft^2 + 18ft^2 + 48ft^2 = 108ft^2

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The answer is 1!!!!!!!!!!!!!!!!!
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2 years ago
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Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

Hence,

The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0
2 years ago
In Quebec, 90 percent of the population subscribes to the Roman Catholic religion. In a random sample of eight Quebecois, find t
AysviL [449]

Answer:

Probability that the sample contains at least five Roman Catholics = 0.995 .

Step-by-step explanation:

We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.

The Binomial distribution probability is given by;

 P(X = r) = \binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,.......

Here, n = number of trials which is 8 in our case

         r = no. of success which is at least 5 in our case

         p = probability of success which is probability of Roman Catholic of

                 0.90 in our case

So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= \binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}

= 56 * 0.9^{5} * (0.1)^{3} + 28 * 0.9^{6} * (0.1)^{2} + 8 * 0.9^{7} * (0.1)^{1} + 1 * 0.9^{8}

= 0.995

Therefore, probability that the sample contains at least five Roman Catholics is 0.995.

3 0
2 years ago
The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio
zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0
2 years ago
The triangular bases of a triangular prism have three congruent sides, each measuring 10 centimeters. The height of each of the
Elina [12.6K]
We know that

[surface area ]=2*[area of the base]+[perimeter of the base]*height
area of the base=10*8.7/2-----> 43.5 cm²
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height of the prism=15 cm
[surface area ]=2*[43.5]+[30]*15------->537 cm²

the answer is
<span>the approximate surface area of the prism is 537 cm</span>²
5 0
2 years ago
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