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Nezavi [6.7K]
3 years ago
5

ANSWER ASAPPP PLEASEEEEEE

Mathematics
1 answer:
Law Incorporation [45]3 years ago
5 0
The following sum is irrational. √16 is rational because it's perfect square, the result is 4. But √8 is not rational because it's not perfect square.

The answer is second option
You might be interested in
Triangle DEF is a right triangle. If FE=5 and DF=13, find DE.
Nadya [2.5K]

Answer:

13

Step-by-step explanation:

I am amusing you are looking for the hypotenuse since this is middle school.

If you have the legs and are looking for the hypotenuse you use the Pythagorean Theorem a^{2} + b^{2} = c^{2}

The legs are a and b, and the hypotenuse is c

The work would look like this

13^{2} + 5^{2} = c^{2} \\169+25= c^{2} \\194= c^{2} \\\sqrt{194} = \sqrt{c^{2} } \\13.9283883... = c

extra notes

  • I rounded to 13
  • If you didn;t know you cancles exponents with roots

so the c^{2} is cancled by aka \sqrt{c}

Hope this helped :)

4 0
3 years ago
A bridge is sketched in the coordinate plane as a parabola represented by the equation h=40-0.01x2, where h refers to the height
ElenaW [278]

Answer:

The length of the bridge is 126.492 feet.

Step-by-step explanation:

Let h(x) = 40-0.01\cdot x^{2}, where x is the position from the middle of the bridge, measured in feet, and h(x) is the height of the bridge at a location of x feet, measured in feet. In this case, the length of the bridge is represented by the distance between the x-intercepts of the parabola, which we now find by factorization:

40-0.01\cdot x^{2} = 0 (Eq. 1)

x^{2} = \frac{40}{0.01}

x =\pm \sqrt{\frac{40}{0.01} }

x = \pm 63.246\,ft

Given that the parabola is symmetrical with respect to y-axis, then the length is two times the magnitude of the value found above, that is:

l = 2\cdot (63.246\,ft)

l = 126.492\,ft

The length of the bridge is 126.492 feet.

3 0
3 years ago
Divide: 5 divided by 5/6
Alenkasestr [34]
I know the answer!!! The answer is 6
3 0
3 years ago
Teacher raises A school system employs teachers at
Cerrena [4.2K]

By adding a constant value to every salary amount, the measures of

central tendency are increased by the amount, while the measures of

dispersion, remains the same

The correct responses are;

(a) <u>The shape of the data remains the same</u>

(b) <u>The mean and median are increased by $1,000</u>

(c) <u>The standard deviation and interquartile range remain the same</u>

Reasons:

The given parameters are;

Present teachers salary = Between $38,000 and $70,000

Amount of raise given to every teacher = $1,000

Required:

Effect of the raise on the following characteristics of the data

(a) Effect on the shape of distribution

The outline shape of the distribution will the same but higher by $1,000

(b) The mean of the data is given as follows;

\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}

Therefore, following an increase of $1,000, we have;

 \overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} =  \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}

\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000

  • Therefore, the new mean, is equal to the initial mean increased by 1,000

Median;

Given that all salaries, x_i, are increased by $1,000, the median salary, x_{med}, is also increased by $1,000

Therefore;

  • The correct response is that the median is increased by $1,000

(c) The standard deviation, σ, is given by \sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x  \right )^{2} }{n}};

Where;

n = The number of teaches;

Given that, we have both a salary, x_i, and the mean, \overline x, increased by $1,000, we can write;

\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x  + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x  - 1000\right )^{2} }{n}}

\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x  - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}

\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x  \right )^{2} }{n}} = \sigma

Therefore;

\sigma_{new} = \sigma; <u>The standard deviation stays the same</u>

Interquartile range;

The interquartile range, IQR = Q₃ - Q₁

New interquartile range, IQR_{new} = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR

Therefore;

  • <u>The interquartile range stays the same</u>

Learn more here:

brainly.com/question/9995782

6 0
3 years ago
Pls help me thanks!!
Leni [432]

Answer:

Its A <3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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