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Vera_Pavlovna [14]
2 years ago
13

Factor 2x² + 13x + 20​

Mathematics
2 answers:
emmasim [6.3K]2 years ago
4 0

Answer:

(2x+5)(x+4)

Step-by-step explanation:

Elena L [17]2 years ago
3 0

9514 1404 393

Answer:

  (x +4)(2x +5)

Step-by-step explanation:

To factor this, you look for factors of 2·20 = 40 that have a sum of 13.

The factor pairs of 40 are ...

  40 = 1·40 = 2·20 = 4·10 = 5·8

The sums of these factor pairs are 41, 22, 14, 13. The last one is the one we want.

There are several ways to use this factor pair. One of the more straightforward methods is to rewrite the middle term as a sum using these numbers:

  2x² +5x +8x +20

  = x(2x +5) +4(2x +5) . . . . . factor pairs of terms

  = (x +4)(2x +5) . . . . . . . . . factor out the common factor

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Step-by-step explanation:

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A medium sized apple has 70 calories. This is 10 calories less than 1\4 of the calories in an old westie chocolate bar.How many
n200080 [17]

Answer:

The answer to your question is: 320

Step-by-step explanation:

Data

1/4 of the calories in a chocolate - 10 = calories in an apple

Equation

    Calories in a chocolate bar = 4(calories in apple + ten)

                                                                    = 4(70 + 10)

                                                                    = 4(80)

                                                                   = 320

                     

7 0
3 years ago
Read 2 more answers
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard dev
Varvara68 [4.7K]

Answer:

The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.

Step-by-step explanation:

We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.

Let \bar X = <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean length of rods = 259.2 cm

           \sigma = standard deviaton = 2.1 cm

           n = sample of steel rods = 17

Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(\bar X > 259 cm)

 

     P(\bar X > 259 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{259-259.2}{\frac{2.1}{\sqrt{17} } } ) = P(Z > -0.39) = P(Z < 0.39)

                                                                = <u>0.65173</u>

The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.

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