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1 year ago

Mr.Anderson paid $19.18 for a CD cover. This amount includes a tax of 8.5%. What was the cost of the CD cover before tax ?

1 answer:

3 0

X * 1.085 = 19.18
solve for x

Why is this how you figure it out? well... you have the price of the CD cover and the total is the price of the CD cover with tax.

Price of the CD cover would be: x * 1 = price of cd cover
Put that tax on there: x * 1.085 = price of cd cover with tax

why .085? move that decimal over two places to the left from the 8.5%

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At the grocery store you can buy a whole chicken that weighs 4 pounds for $3.88. You can also buy a package of chicken drumstick

kirza4 [7]

Whole chicken unit price= $0.97
Chicken drumstick unit price= $1.02
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2 years ago

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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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2 years ago

Kent has a collection of pennies and nickels with a value of 1.98$. the number of pennies he has is five less than twice the num

Alex73 [517]

<span>Kent has a collection of pennies and nickles with a value of $1.98. The number of pennies he has is five less than twice the number of nickles. How many of each coin does Kent have?
----
p + 5n = 198
p = 2n - 5
</span>---
Sub for p
2n-5 + 5n = 198

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1 year ago

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What is 200 percent of (0.020(5/4) + 3 ((1/5) - (1/4)))

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Answer:

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Step-by-step explanation:

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2 years ago

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mrs_skeptik [129]

Question Completion:

How are the percentages distributed? Is the distribution skewed? Are there any gaps? (Select all that apply.)

Answer:

1. The percentages are concentrated from 20% to 60%.

2. These data are strongly skewed left.

3. There are no gaps in the data.

Step-by-step explanation:

1. Data

Percentage loss of wetlands per state

46 37 36 42 81 20 73 59 35 50

87 52 24 27 38 56 39 74 56 31

27 91 46 9 54 52 30 33 28 35

35 23 90 72 85 42 59 50 49

48 38 60 46 87 50 89 49 67

2. Re-arrangement of

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9 20 23 24 27 27 28 30

31 33 35 35 35 36 37 38

38 39 42 42 46 46 46 48

49 49 50 50 50 52 52 54

56 56 59 59 60 67 72 73

74 81 85 87 87 89 90 91

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