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3 years ago

Solve this equation: 80 = 3y + 2y + 4 + 1.

2 answers:

6 0

Let us first write down the given equation.
80 = 3y + 2y + 4 + 1
80 = 5y + 5
80 - 5 = 5y
75 = 5y
Just reversing both sides we get
5y = 75
Now dividing both sides by 5 we get
y = 15
So the value of y is 15. I hope the procedure for solving the problem is clear to you. In future you can solve such problems with ease following the procedure described.

Pavel [41]3 years ago

6 0

$80 = 3y + 2y + 4 + 1 \\ \\ 80 = 5y + 4 + 1 \\ \\ 80 = 5y + 5 \\ \\ 80 - 5 = 5y \\ \\ 75 = 5y \\ \\ \frac{75}{5} = y \\ \\ 15 = y \\ \\ y = 15 \\ \\ Answer: \fbox {y = 15}$

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Answer:

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Jo's house is located at (-4,-2), Elenas house is a reflection of Jo's house across the x-axis. What is the coordinate of Elenas

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3 years ago

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The ratio of pencils to pens in the drawer is 8:14. If there are actually twice as many pencils as in the ratio, how many pencil

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let the number of pencils and pens be 8x and 14x

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4 years ago

The side length of a cube is (x^2-1/2). Determine the volume of the cube.

stellarik [79]

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Step-by-step explanation:

x^6 − (3x^4)/2 + (3x^2)/4 − 1/8

The volume of a cube is Length^3.

If a length is (x^2-1/2, then cube root it

It will then be (x^2-1/2) x (x^2-1/2) x (x^2-1/2).

FOIL each term and evaluvate

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3 years ago

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A cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled. Each box of cereal is supposed to

Vaselesa [24]

Answer:

We conclude that the population average weight of a cereal box is equal to 13 ounces.

Step-by-step explanation:

We are given that a cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled.

A random sample of 36 boxes is tested. The sample average weight is 12.85 ounces and the sample standard deviation is 0.75 ounces.

Let $\mu$ = population average weight of a cereal box

So, Null Hypothesis, $H_0$ : $\mu$ = 13 ounces {means that the population average weight of a cereal box is equal to 13 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 13 ounces {means that the population average weight of a cereal box differs from 13 ounces}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

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

where, $\bar X$ = sample average weight = 12.85 ounces

$\sigma$ = sample standard deviation = 0.75 ounces

n = sample of boxes = 36

So, test statistics = $\frac{12.85-13}{\frac{0.75}{\sqrt{36} } }$ ~ $t_3_5$

= -1.20

The value of the test statistics is -1.20.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the t table gives critical values between -2.03 and 2.03 at 35 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population average weight of a cereal box is equal to 13 ounces.

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3 years ago