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AURORKA [14]
2 years ago
8

Please help me. I need to know the answer.

Mathematics
1 answer:
zavuch27 [327]2 years ago
6 0

Step-by-step explanation:

Given =

AC = 6cm

BC = 9cm (hyp)

AB =

By pyth

A^2 = B^2 + C^2

9 = 6 + AB

AB = 9-6 = 3

AB =3

Hope this helps

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Sandy bought 3/4 yard of ribbonand 2/3 yard of white ribbon how much ribbon did she but altogether
Neko [114]
To answer this you need to find the LCD (least common denominator) 2/3times4=8/12 and 3/4times3=9/12 and 9+8=17 then 17-12=5 so she bought 1 5/12 yards of ribbon
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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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