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3 years ago

8. a number increased by five and then divided by two

Mathematics

2 answers:

ivanzaharov [21]3 years ago

5 0

Answer:

Step-by-step explanation:

Butoxors [25]3 years ago

4 0

Answer:3

Step-by-step explanation:

1+5=6

6/2=3

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What is differnce when there is parentheses around a number being squared?

Svetach [21]

When parentheses are around a number being squared, the output value will always be positive.

For example: -2² vs (-2)²
-2² means the negative of the square of two, or -4
(-2)² means the square of negative two, or 4

5 0

3 years ago

What is the opposite of -21? show all work

bogdanovich [222]

21 is the opposite of -21

8 0

3 years ago

Two cables support a 800-lb weight, as shown. Find the tension in each cable.

jeka94

Answer:

892 lb (right)
653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

Rcos(45°) +Lcos(75°) = 800

Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

= 800sin(75°)/sin(120°) ≈ 892 . . . pounds

L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

(right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0

3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

3 years ago

A museum curator is hanging 7 paintings in a row for an exhibit. There are 4 Renaissance paintings and 3 Baroque paintings. From

Marina CMI [18]

Answer:

144 ways

Step-by-step explanation:

Number of paintings = 7

Renaissance = 4

Baroque = 3

We are hanging from left to right and we will first hang Renaissance painting before baroque painting.

For Renaissance we have 4! Ways of doing so. 4 x3x2x1 = 24

For baroque we have 3! Ways of doing so. 3x2x1 = 6

We have 4!ways x 3!ways

= (4x3x2x1) * (3x2x1) ways

= 144 ways

Therefore we have 144 ways to hang the painting.

6 0

3 years ago