The price of a CD that sells for 21% more than<br> the amount (m) needed to manufacture the CD

A candy maker buys a bar of chocolate weighing 162 oz. About how many lb does the bar weigh?

Natali [406]

Use dimensional analysis. 1 lb = 16 oz will be the conversion factor. 162 oz x (1 lb/16 oz) = 10.125 lb.

7 0

3 years ago

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Solve using quadratic formula<br> 6x^2-x=2

Nadya [2.5K]

$6x^2-x=2\implies 6x^2-1x-2=0 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{6}x^2\stackrel{\stackrel{b}{\downarrow }}{-1}x\stackrel{\stackrel{c}{\downarrow }}{-2}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$x = \cfrac{ -(-1) \pm \sqrt { (-1)^2 -4(6)(-2)}}{2(6)}\implies x = \cfrac{1\pm\sqrt{1+48}}{12} \\\\\\ x = \cfrac{1\pm\sqrt{49}}{12}\implies x = \cfrac{1\pm 7}{12}\implies x = \begin{cases} \frac{8}{12}\to &\frac{2}{3}\\[1em] -\frac{6}{12}\to &-\frac{1}{2} \end{cases}$

7 0

3 years ago

Alejandro bought a new refrigerator for $856 with a 25% discount and

oksano4ka [1.4K]

Answer:

$695.94

Step-by-step explanation:

25% of 856 = 214

6.3% of 856 = 53.928

856 + 53.928 = 909.928

909.928 - 214 = 695.928 about 695.94

7 0

3 years ago

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Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

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

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

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

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

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

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago

Help ASAP, i don't understand this.

Inessa [10]

The given polynomial expressed as partial fraction is 5x^3 + 9/x

<h3>Polynomial function and partial fractions</h3>

Polynomial function are function that have a leading degree of 3. Given the expression below;

f(x) = (5x^4+9)/x

Applying partial fraction

Since the part fraction means separating the fraction into two parts, hence;

f(x) = 5x^4/x + 9/x

Simplify

Since there is value of x at both numerator and denominator, it will cancel out to have

f(x) = 5x^3 + 9/x

From the result, we can see that p(x) = 5x^3 and k = 9

Hence the given polynomial expressed as partial fraction is 5x^3 + 9/x

Learn more on polynomial here: brainly.com/question/2833285

#SPJ1

8 0

2 years ago

The price of a CD that sells for 21% more than the amount (m) needed to manufacture the CD