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Tom [10]
3 years ago
9

I need help with solving this quadratic equation by factoring. the problem I need help with is number 5.

Mathematics
1 answer:
otez555 [7]3 years ago
4 0
I think it is - - (4/3), 3/2
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What two numbers add to 12 and multiply to 35
inessss [21]
5 and 7
5 + 7 = 12
5 x 7 = 35
5 0
3 years ago
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Lisa is 10 cm taller than her friend ian. Ian is 14 cm taller than jim. Every month, their heights increase by 2 cm. In 7 months
koban [17]

Answer:  Ian's height = x+14 = 69+14 = 83 cm

Explanation :

Let the height of Jim be x

Let the  height of Ian be x+4

Let the height of Lisa be x+4+10=x+14

Since their heights increase by 2cm every months

so after 7 months ,

Jim's  height = x+14

Ian's  height= x+4+14=x+18

Lisa's  height= x+14+14=x+28

According to question,

Sum of Ian's and Jim's height = 170cm

x+18+x+14 = 170\\2x+32 =170\\2x =170-32\\2x = 138

⇒x = \frac{138}{2}

⇒ x= 69cm

∴ Ian's height = x+14 = 69+14 = 83 cm

5 0
2 years ago
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I don’t understand these question
Lemur [1.5K]
I believe question 3 is B and question 4 is B
5 0
2 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

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

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

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

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0
2 years ago
A proof should always begin with stating the given information.<br> True<br> False
const2013 [10]
<span>A proof should always begin with stating the given information.
True</span>
5 0
3 years ago
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