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garik1379 [7]
3 years ago
10

Here is a triangular prism. 4 cm 10 cm 5cm 5cm 6cm

Mathematics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

(A)  120 cm³

(B) 184 cm²

Step-by-step explanation:

(A)

Assuming the diagram attached is thesame as the one needed to complete the question,

Applying

V = A'×L................... Equation 1

Where V = volume of the prism, A' = Area of the triangular base, L = Length of the prism.

But,

A' = 1/2bh............. Equation 2

Where b = base of the triangle, h = height of the triangle.

Substitute equation 2 into equation 1

V = 1/2bh(L)............... Equation 3

From the question,

Given: b = 6 cm, h = 4 cm, L = 10 cm

Substitute these values into equation 3

V = 1/2(6×4×10)

V = 120 cm³

(B)

Surface area of the prism(A) = bh+L(a+b+c).............. Equation 4

Given: b = 6 cm, b = 5 cm, c = 5 cm, L = 10 cm, h = 4 cm

Substitute these values into equation 4

A = (6×4)+10(5+5+6)

A = 24+160

A = 184 cm²

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Put these in order starting with the smallest.<br>9 3/2 <br>27 1/3 <br>125 2/3​
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Answer:

second, third,first

Step-by-step explanation:

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3/2=1.5

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3 years ago
A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

In this question, x = 5 and \sigma = 2. The equation becomes

\displaystyle z = \frac{5 - \mu}{2}.

To solve for \mu, the mean of this distribution, the only thing that needs to be found is the value of z. Since

The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

The problem is that the normal distribution tables list only the value of P(0 \le Z \le z) for z \ge 0. To estimate  z from P(Z \le z) = 0.6985, it would be necessary to find the appropriate

Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

Therefore:

\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}.

Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

  • z = 0.52,
  • x =5, and
  • \sigma = 2,

Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

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3 years ago
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