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3 years ago

7 1/7-3 2/7 not as a improper fraction

Mathematics

1 answer:

lapo4ka [179]3 years ago

3 0

Answer:

6 6/7

Step-by-step explanation:

Hope this helps:)

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Your parents took your family out to breakfast. Your parents wanted to give the waiter a 15% tip. If the total amount of the bre

zepelin [54]

Answer:

$6.30 for a tip

Step-by-step explanation:

%48.30 as a total

4 0

3 years ago

The wares want to buy and new computer. The regular price is $1,049. The store is offering a 20% discount and a sales tax of 5.2

kozerog [31]

883.26 dollars is what I got.

7 0

4 years ago

-1/6 + 2/3 ( 9 + - 3/4) + - 1/2

Liono4ka [1.6K]

First simplify the section in the parenthesis.
-1/6 + 2/3(8 1/4) + -1/2
Then multiply 2/3 by 8 1/4.
-1/6 + 5 1/2 + -1/2
Add -1/2 to 5 1/2.
-1/6 + 5
Add 5 to -1/6.
4 5/6 is the fully simplified answer.
Hope this helps!

7 0

3 years ago

A company manufactures a brand of lightbulb with a lifetime in months that is normally distributed with mean 3 and variance 1. A

Ratling [72]

Answer:

The smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772 is 16.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

n values from a normal distribution:

The mean is $\mu n$ and the standard deviation is $s = \sigma\sqrt{n}$

A company manufactures a brand of lightbulb with a lifetime in months that is normally distributed with mean 3 and variance 1.

This means that $\mu = 3, \sigma = \sqrt{1} = 1$

For n bulbs:

The distribution for the sum of n bulds has $\mu = 3n, \sigma = \sqrt{n}$

What is the smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772?

We want that: $S_{n} \geq 40 = 0.9772$ .

This means that when X = 40, Z has a pvalue of 1 - 0.9772 = 0.0228, that is, when $X = 40, Z = -2$ . So

$Z = \frac{X - \mu}{\sigma}$

$-2 = \frac{40 - 3n}{\sqrt{n}}$

$-2\sqrt{n} = 40 - 3n$

$3n - 2\sqrt{n} - 40 = 0$

Using $y = \sqrt{n}$

$3y^2 - 2y - 40 = 0$

Which is a quadratic equation with $a = 3, b = -2, y = -40$

$\Delta = b^{2} - 4ac = (-2)^2 - 4(3)(-40) = 484$

$y_{1} = \frac{-(-2) + \sqrt{484}}{2*3} = 4$

$y_{2} = \frac{-(-2) - \sqrt{484}}{2*3} = -...$

Since y and n have both to be positive:

$y = \sqrt{n}$

$\sqrt{n} = 4$

$(\sqrt{n})^2 = 4^2$

$n = 16$

The smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772 is 16.

4 0

3 years ago

just a $50 in her savings accountl she then strated adding $50 each week to her account for the next 5 weeks

just olya [345]

Answer:300

Step-by-step explanation:

50+5x=?

Literally just multiply it by 50 by 5 which would be 250 plus what she already had in her account 250+50= 300

5 0

3 years ago

Read 2 more answers