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2 years ago

Juanita has saved 30% of the money that she needs to buy a new bicycle.

1 answer:

3 0

well, the bicycle costs "x" which is the 100% of its price, we know she has 30% thus far, and we also know that that is 66 bucks.

$\begin{array}{ccll} amount&\%\\ \cline{1-2} x&100\\ 66&30 \end{array}\implies \cfrac{x}{66}=\cfrac{100}{30}\implies \cfrac{x}{66}=\cfrac{10}{3} \\\\\\ 3x=660\implies x\cfrac{660}{3}\implies x=220$

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Solve for x. (16x + 23)°

erma4kov [3.2K]

16x +23
-16 -16
X = 7....................

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2 years ago

2-1 help anyone pls............

ValentinkaMS [17]

Answer:

In what book?

Step-by-step explanation:

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3 years ago

Put these numbers in order from greatest the least 28/13 -0.4 1 6/20

tiny-mole [99]

Answer:

Greatest to least

28/13, 1, 6/20, -0.4

Step-by-step explanation:

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2 years ago

In a telephone poll, 18 people said they like shopping and 27 people said they do not like shopping. What is the ratio of the nu

Rainbow [258]

Answer:

2/3

Step-by-step explanation:

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2 years ago

There are 10 employees in a particular division of a company. Their salaries have a mean of $70,000, a median of $55,000, and a

alekssr [168]

Answer:

A) $bar{x}_{new}=160,000$

B) Median remains the same.

C) $sigma_{new}=300998.34$

Step-by-step explanation:

Consider the complete question attached below.

No. of employees = n = 10

Given mean = $70,000

Median = $55,000

Standard deviation = $60,000

Largest number on the list = $100,000

Accidentally changed to = $1,000,000

Modified mean, media, SD =?

A) Modified Mean:

$\bar{x}=\frac{\sum x}{n} = 70,000\\\\\sum x=(\bar{x})(n) = (70,000)(10)\\\\\sum x=700,000\\\\\sum x_{new} =700,000 -100,000+1,000,000\\\\\sum x_{new}=1,600,000\\\\\bar{x}_{new}=\frac{1,600,000}{10}\\\\\bar{x}_{new}=160,000$

B) Modified Median:

Median remains same and is not affected by changing highest value.

C) Modified SD:

Standard deviation is given by formula:

$\sigma=\sqrt{\frac{\sum x^{2}-n\bar{x}}{N-1}}---(1)\\\\\sigma_{new}=\sqrt{\frac{\sum x_{new}^{2}-n\bar{x}_{new}}{N-1}}---(2)\\\\From\,\, (1)\\\\\sum x^{2}=(N-1)\sigma^{2}+n\bar{x}$

$\sum x^{2}=(10-1)(60,000)^{2}+(10)(70,000)^{2}\\\\\sum x^{2}=8.14\times 10^{10}\\\\\sum x_{new}^{2}= 8.14\times 10^{10}-(10,0000)^{2}+(1,000,000)^{2}\\\\\sum x_{new}^{2}=1.0714\times 10^{12}\\\\Using\,\, (2)\\\sigma_{new}=\sqrt{\frac{1}{9}(1.0714\times 10^{12}-(10)(1.6\times 10^{5})}\\\\\sigma_{new}=\sqrt{9.06\times 10^{11}}\\\\\sigma_{new}=300998.34$

7 0

3 years ago