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3 years ago

Help on this easy pre-algebra problem please!! ASAP

Mathematics

2 answers:

Aleks [24]3 years ago

5 0

About 7
because 5^2+5^2 is 50 and then you take the sq root of 50 which is about 7

Lana71 [14]3 years ago

5 0

The answer is (d) 50

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Rebecca is making bracelets to sell at a charity event. Each bracelet takes 1 1/4 of a foot of embroidery floss to make.If Rebec

Ivan

Answer:

Step-by-step explanation:

She has 15 feet of floss. 15/1.25 is 12.

Hope this helps!!

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2 years ago

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{67, 68, 68, 71, 73, 73, 73, 74, 74, 74, 74, 75, 75, 77, 78, 79, 79, 80, 80}

Fynjy0 [20]

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Step-by-step explanation:

Mode is the number that shows up the most in the data.

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2 years ago

A limited edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.

RideAnS [48]

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3 years ago

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Maru [420]

Answer:

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:

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

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

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3 years ago