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disa [49]
3 years ago
5

-X + 5 + 3x = 4 + 2x + 1

Mathematics
1 answer:
zloy xaker [14]3 years ago
5 0

Answer;

its answer is 0.

Step-by-step explanation:

  1. -X+5+3X=4+2X+1=
  2. -X+3X-2X+5-4-1=
  3. -3X+3X+5-5=
  4. cut -3X+3Xand +5-5=
  5. answer is 0
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Rodney buys 6 pounds of bananas for 5.10$. how much do bananas cost?
emmasim [6.3K]

Answer:

  $0.85 per pound

Step-by-step explanation:

The cost in dollars per pound is found by dividing dollars by pounds:

  $5.10/(6 lb) = $0.85 /lb

Bananas cost $0.85 per pound.

5 0
3 years ago
A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s
Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.          

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

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

We have to find the value of x such that the probability is 0.0793.

P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793  

= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793  

=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207  

Calculation the value from standard normal z table, we have,  

P(z \leq 1.410) = 0.9207

\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51  

Hence, 88.51 is the minimum score needed to receive a grade of A.

3 0
3 years ago
Please help me with my homework
geniusboy [140]

Answer:

A. 5

B. 6

I hope this is helpful for you! Unless my calculator is wrong!

7 0
2 years ago
A diver dove to a location 63/5 meters below sea level .he then dove to a second location 8 1/5 meters below sea level how many
satela [25.4K]
You just take the difference between them and the answer is 1 3/5
3 0
3 years ago
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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