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Alisiya [41]
3 years ago
7

Please help click on pictures ​

Mathematics
1 answer:
shutvik [7]3 years ago
3 0

Answer:

The area of the triangle is 40cm^2

Step-by-step explanation:

Calculate the area of a triangle using the equation: 1/2hb

Look at the attached picture and the equation should make sense.

You can use 8cm and 10cm.

8*10 = 80

1/2(80) = 40

So, the area of the triangle is 40cm^2

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❤ <em>Hey There!! ❤</em>

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Step-by-step explanation:

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The height, in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84. (
Schach [20]

Answer:

Step-by-step explanation:

Given that the height in  inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84.

X is N(64, 2.8)

Or Z = \frac{x-64}{2.8}

a)  the probability that the height of a randomly chosen woman is between 59.8 and 68.2 inches.

=P(59.8

b) P(X\geq 59)\\= P(X\geq -1.78)\\ \\=0.9625

c) For 4 women to be height 260 inches is equivalent to

4x will be normal with mean (64*4) and std dev (2.8*4)

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3 years ago
The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2
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Answer:

a) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 0 are x_{1}\approx -0.25 and x_{2} \approx 1.25.

b) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 2 are x_{1}\approx -0.5 and x_{2} \approx 1.5

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial y = 4\cdot x^{2}-4\cdot x -1, where x is related to the horizontal axis of the Cartesian plane, whereas y is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:

a) 4\cdot x^{2}-4\cdot x -1 = 0

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.25, x_{2} \approx 1.25

b) 4\cdot x^{2}-4\cdot x -1 = 2

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.5, x_{2} \approx 1.5

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