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3 years ago

Why would negative numbers not make sense for the values of the variables

2 answers:

7 0

One reason that negative numbers would not make sense for values of the variables is that the solution having a negative value is not valid. It might be that the expression represents a real word problem and a negative value will not make any sense. Hope this helps.

atroni [7]3 years ago

5 0

For this case, what we are going to do is answer the question using an example.

Suppose that the equation of the height of a ball as a function of time is given by:

$h (t) = -16t ^ 2 + 40t + 100$

We want to find the time when the ball reaches the ground.

To do this, we must match the equation to zero:

$-16t ^ 2 + 40t + 100 = 0$

From here, we clear the time.

The polynomial is of the second degree.

Therefore, it has two solutions.

Since we want to find time, then negative solutions do not make sense in this case.

Answer:

Negative values do not make sense for variables in some real-life problems.

Example: find the time when a ball hits the ground.

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How many 1/3 cup servings of juice are in a 5 cup container

vesna_86 [32]

Answer:

We can get 15 (1/3 cup) servings with a 5 cup container

Step-by-step explanation:

To solve this problem, we can easily divide 5 by (1/3) to find how many servings can be obtained

5 / (1/3)

= 5 / (0.3333)

= 15

This can be checked by adding together (1/3) fifteen times

(1/3) + (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)+ (1/3)

= 15/3 = 5

7 0

3 years ago

Find solution to the system of linear equations. 5x1 + x2 = 0 , 25x1 + 5x2 = 0

Zigmanuir [339]

Answer:

the system of equation has infinite solution

Step-by-step explanation:

$5x_1 + x_2 = 0$ , $25x_1 + 5x_2 = 0$

Solve the first equation for x_2

Subtract 5x1 on both sides

$5x_1 + x_2 = 0$

$x_2 =-5x_1$

Now substitute -5x1 on second equation

$25x_1 + 5x_2 = 0$

$25x_1 + 5(-5x_1)= 0$

$25x_1-25x_1= 0$

0=0

So the system of equation has infinite solution

5 0

3 years ago

Read 2 more answers

Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = mean sunspot activity during the Spanish colonial period.

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, the test statistics = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

(c) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = 0.1401

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, 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 mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0

4 years ago

Factor the trinomial below. x2 + 14x + 48

elixir [45]

IF IT EQUALS 0
1. find two number that multiplied to 48 and adds to 14, which are 6 and 8.

2. substitute the new numbers in with x to get x^2 + 6x + 8x + 48.

3. factor out the x and the 8 to get x(x+6)+8(x+6).

4. x = -6, x = -8

IF IT DOES NOT EQUAL 0
then (x+6)*(x+8) is your answer.

4 0

4 years ago

Help me to do this question

gtnhenbr [62]

Step-by-step explanation:

here is tha answer..hope it helps..mark me as brainliest

4 0

3 years ago