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2 years ago

The high temperatures for the last seven days are: High Temperatures: 81, 78, 83, 89, 80, 87, 78

Mathematics

2 answers:

slava [35]2 years ago

8 0

Answer: 78

Step-by-step explanation:

The mode is the number that appears the most number of times. In this case the number 78 appears twice unlike the other numbers that only appear once. Hence, 78 is the mode :)

Semenov [28]2 years ago

7 0

Answer:

Step-by-step explanation:

The mode is the value that appears most often in a set of data. The mode of a discrete probability distribution is the value x at which its probability mass function takes its maximum value. In other words, it is the value that is most likely to be sampled.

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Quation that show an estimate of 7,203+299

Dmitry [639]

The two numbers are very near 7200 and 300. The sum of those is a good estimate of your sum:

... 7200 + 300 = 7500

_____

If you're taught methods of addition that include "make a ten" or "make a hundred", you know you can "borrow" a 1 from 7203 (making it 7202) and loan that 1 to 299 to make it 300. Now, you know your sum is exactly 7202+300 = 7502, with no estimation necessary.

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2 years ago

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Your gym membership costs $33 per month after an initial membership fee . you paid a total of $228 after 6 months.

Lubov Fominskaja [6]

So, what I did was multiply 33*6 and got 198 and then 228-198 to find out that the initial fee was 30 dollars. then multiply 33*9+30 to get a 9 month total cost of $327. Hope from here you are able to set up your equation now you have all the information needed :)

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3 years ago

How many terms are there in this expression? 3a + b - 4c ——- 3 6 5

Dafna1 [17]

There's 3 terms

3a-----1

+b -----2

-4c------3

3 0

2 years ago

Explain how you can write 3/2 as the product of a whole number and a unit fraction.

Nitella [24]

Answer:

Step-by-step explanation:

When 3 is divided by 2, the quotient is 1 and remainder is 1

Whole number is the quotient.

Fraction : remainder

divisor

$1\frac{1}{2}$

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3 years ago

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Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

3 years ago