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3 years ago

2x + 3 + 4x + 6 + 12x + 18 = 477 Please answer with solving steps. Thanks guys :)

2 answers:

4 0

yoi have to add the variables so 2x +4x+12x=18x and concombine like terms so 3 +6+18=27 so you have to put tbat together 18x+27=477 then you have to get x alone so you have to subtract 27 in both sides 27-27=0 then 477-27=450 then you have to do the oppsite of multiply so you have to divide 18x/18x and 450/18x 18÷18 is 1 so you just have to put x and then divide 450÷18=25 so x=25.

dybincka [34]3 years ago

3 0

Answer:

x= 25

Step-by-step explanation:

2x + 3 + 4x + 6 + 12x + 18 = 477

FIRST combine like terms

take all "x"

2+4+12= 18x

take all whole numbers

3+6+18= 27

18x+27= 477

now solve for x

-27 on both sides

18x= 450

÷18 on both sides

x= 25

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It is given in the question that:-

Richard fills his pool at a rate of 6 gallons per minute.

Jason's pool already contains 200 gallons, and he fills it at a rate of 2 gallons per minute.

After 20 minutes the amount of liquid in the tank will be:-

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For Jason= 2 gallons per minute x 20 =40 gallons

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2 years ago

Which of the following expressions are equivalent to -5/-7×-11/-6

USPshnik [31]

Answer:

Step-by-step explanation:

-5/-7×-11/-6

= (-5 * -11) / (-7*-6)

= 55 / 42.

A. = - (6/-11) = 6/11

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B. - (-11/7) = 11/7

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3 years ago

What are the possible values of x and y for two distinct points, (5, –2) and (x, y), to represent a function?

bonufazy [111]

The correct answer is:

Any real number for x except 5, and any real number for y.

Explanation:

A function is a relation in which each element of the domain, or x-value, is mapped to one element of the range, or y-value. This means that no x can be mapped to more than one y, so if we have the same number used twice for x, we do not have a function.

This means that 5 cannot be used for x in the other ordered pair.

Since there is no restriction on y in order to be a function, y can be any real number.

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3 years ago

Read 2 more answers

A bowl has the shape of a hemisphere whose diameter is 14 centimeters. If this bowl is filled with water, how many cubic centime

ASHA 777 [7]

Answer:

The cubic centimeters of water it will take to fill the bowl is 718.67 cm³

Step-by-step explanation:

Given;

diameter of the hemisphere, d = 14 cm

radius of the hemisphere, r = d/2 = 14/2 = 7 cm

The volume of the hemisphere is calculated as;

$V = \frac{2}{3} \pi r^3\\\\V = \frac{2}{3} \times \frac{22}{7} \times (7 \ cm)^3\\\\V = 718.67 \ cm^3$

The cubic centimeters of water it will take to fill the bowl is 718.67 cm³

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3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

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3 years ago