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3 years ago

Evaluate each expression if A=Z, B=5, X=4, and N=10. what is the answer for [a+8 (b-2)]^2÷4

Mathematics

1 answer:

nalin [4]3 years ago

8 0

Hi!

<h3>Put in the values. </h3>

[z+8(5-2)]^2/4

<h3>Now use PEMDAS to solve. </h3>

Parentheses - Subtraction

[z+8*3]^2/4

Parentheses - Multiplication

[z+24]^2/4

Exponents

z+576/4

Division

<h2>When you evaluate [a+8 (b-2)]^2÷4, you get z + 144</h2>

Hope this helps! :)

-Peredhel

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On Monday the price of one share of Webb company stock was 24.85. By Friday the price was 25.004. By how much did the price chan

ella [17]

Answer:

Change in price = 0.154

Step-by-step explanation:

Given that,

The price on Monday = 24.85

The price on Friday = 25.004

We need to find the change in price from Monday to Friday. It can be calculated as follows :

Change = 25.004 - 24.85

= 0.154

So, the required change in the price is equal to 0.154 .

8 0

3 years ago

Winona’s bowling scores for the past nine weeks are 195, 180, 195, 212, 208, 231, 179, 246, and 195. Find the mean, median, and

belka [17]

Answer: Mean = 204.6 , Median = 195 and mode =195

Step-by-step explanation:

The given data for Winona’s bowling scores for the past nine weeks : 195, 180, 195, 212, 208, 231, 179, 246, 195.

The formula to find Mean is given by :-

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}\\\\=\dfrac{1841}{9}=204.555555556\approx204.6$

For median , first we arrange data values in order as

179, 180, 195, 195, 195, 208, 212, 231, 246

Since the number of data values is 9 (odd).

∴ Median = Middle most data value = 195

Also, Mode = Most occurring data value = 195

3 0

3 years ago

What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

jasenka [17]

<u>Answer:</u>

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

<u>Solution: </u>

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$