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serious [3.7K]
3 years ago
8

3c-d=T Could you solve for c

Mathematics
2 answers:
RSB [31]3 years ago
7 0
- - Let's solve for C.

Starting off with our equation:
3c+d=T

Transform:
3c=T+D

Solve for that:
\frac{3c}{3} = \frac{T+D}{3}


Solve for that for the last time and your answer is \frac{T+D}{3} =C



Nady [450]3 years ago
3 0
Yes, we can solve for c.

3c - d=T
     +d   +d

3c = t+d

3c/3 = \frac{t+d}{3}

Answer:  c =  \frac{t + d}{3}
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Simplify. (√5+4) (√5 - 2)​
son4ous [18]

Answer:

2√5 - 3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Expand by FOIL

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

(√5 + 4)(√5 - 2)

<u>Step 2: Simplify</u>

  1. Expand [FOIL]:                                                                                                  (√5)² - 2√5 + 4√5 - 8
  2. Combine like terms:                                                                                           (√5)² + 2√5 - 8
  3. Evaluate exponents:                                                                                        5 + 2√5 - 8
  4. Combine like terms:                                                                                         2√5 - 3
6 0
2 years ago
Read 2 more answers
Which scenario best matches the inequality x&gt;3?
jek_recluse [69]
The correct answer is B.
8 0
2 years ago
Read 2 more answers
The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day
Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(between 236 and 281 days)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%

b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

1-\dfrac{1}{(2)^2} = 75\%

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0
3 years ago
12. (07.06 LC) A library building is in the shape of a rectangle. Its floor has a length of (3x + 5) meters and a width of (5x −
Goshia [24]

ANSWER

15 {x}^{2} + 22x - 5

EXPLANATION

It was given that, the length of the rectangular building is

(3x + 5) \: meters

and the width of the is

(5x - 1) \: meters

The area of a rectangular building is calculated using the formula for finding the area of a rectangle.

A = l \times w

Since the dimensions are given in terms of x, the area is also a function of x,

A(x) = (3x +5 )(5x - 1)

We expand to get,

A(x) = 3x(5x - 1) + 5(5x - 1)

A(x) = 15 {x}^{2}  - 3x + 25x - 5

A(x) = 15 {x}^{2} + 22x - 5

meters square

8 0
3 years ago
Which real numbers are zeros of the function?
Galina-37 [17]

Answer: 0, 1/3, -3, 3

Explanation:

I got this question right on k12.

<h2>sincerely, The greatest </h2>

7 0
3 years ago
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