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3 years ago

-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:

-X+5+3X=4+2X+1=
-X+3X-2X+5-4-1=
-3X+3X+5-5=
cut -3X+3Xand +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.

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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.

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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!

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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

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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:

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

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

$q^0=1$

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

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

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

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

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

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

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3 years ago