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

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

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

3 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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