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3 years ago

(8n+1)(6n-3) please solve in quadratic formula

Mathematics

1 answer:

cestrela7 [59]3 years ago

5 0

Use the distributive property.

$(8n+1)(6n-3)=(8n)(6n)+(8n)(-3)+(1)(6n)+(1)(-3)\\\\=48n^2-24n+6n-3=48n^2-18n-3$

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How can you analyze connections between linear equations and use them to solve problems

Alexeev081 [22]

Answer:

Step-by-step explanation:

One of the more obvious "connections" between linear equations is the presence of the same two variables (e. g., x and y) in these equations.

Assuming that your two equations are distinct (neither is merely a multiple of the other), we can use the "elimination by addition and subtraction" method to eliminate one variable, leaving us with an equation in one variable, solve this 1-variable (e. g., in x) equation, and then use the resulting value in the other equation to find the value of the other variable (e. g., y). By doing this we find a unique solution (a, b) that satisfies both original equations. Not only that, but also this solution (a, b) will also satisfy both of the original linear equations.

I urge you to think about what you mean by "analyze connections."

6 0

3 years ago

The lassos high school senior class raised funds for an end of the year cruise getaway. The city gave them a special package for

kirill115 [55]

Answer:

Step-by-step explanation:

x = total number of students

1500 + 367x = 16,600

367x = 16,600 - 1500

367x = 15100

x = 15100 / 367

x = 41.14.......rounds to 41 students <==

this only works if the port fees and taxes that the city gave them are included in the money the class raised

5 0

3 years ago

A manufacturer must test that his bolts are 4.00 cm long when they come off the assembly line. He must recalibrate his machines

storchak [24]

Answer:

The null hypothesis is $H_o: \mu = 4$ .

The alternate hypothesis is $H_a: \mu \neq 4$

The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.

Step-by-step explanation:

A manufacturer must test that his bolts are 4.00 cm long when they come off the assembly line.

This means that the null hypothesis is that they are 4.00 cm long, that is:

$H_o: \mu = 4$

At the alternate hypothesis we test if it is different. So

$H_a: \mu \neq 4$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

4 is tested at the null hypothesis:

This means that $\mu = 4$

After sampling 196 randomly selected bolts off the assembly line, he calculates the sample mean to be 4.14 cm.

This means that $n = 196, X = 4.14$

He knows that the population standard deviation is 0.83 cm.

This means that $\sigma = 0.83$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{4.14 - 4}{\frac{0.83}{\sqrt{196}}}$

$z = 2.36$

Pvalue of the test and decision:

The pvalue of the test is the probability that the sample mean is different by at least 4.14 - 4 = 0.14 from the target value, which is P(|z| > 2.36), which is 2 multiplied by the pvalue of z = -2.36.

Looking at the z-table, z = -2.36 has a pvalue of 0.0091

2*0.0091 = 0.0182

The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.

5 0

3 years ago

Which will result in a difference of squares?

Vedmedyk [2.9K]

Please show the choices

8 0

4 years ago

Forty-two of the students in band own their instrument.

goldfiish [28.3K]

Answer:

112 students

Step-by-step explanation:

Let T be the total number of students in the band.

From the question,

We were told that 37.5% of the total owns their individual instruments and this amount to 42 students

The total number of students can calculated for as follows:

37.5% x T = 42

37.5/100 x T = 42

Cross multiply to express in linear form

37.5 x T = 100 x 42

Divide both side by 37.5

T = (100 x 42) /37.5

T = 112

Therefore, there are 112 students in the band

6 0

3 years ago