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GalinKa [24]

2 years ago

middle" class="latex-formula">

Mathematics

2 answers:

Ostrovityanka [42]2 years ago

5 0

You can subtract a negative number from a positive number or add a positive number to a negative number

Example:

4 - (7) = -3

Or

-8 + 9 = 1

Whatever you are adding has to be greater than the number you are adding on too

Example:

-3 + 5 = 2
Notice how the five is greater than 3

Whatever you are subtracting has to be Greater than the number you are subtracting from.

Example:

5 - 8 = -3
Notice how the eight is greater than five.

The situation changed when adding or subtracting with negative numbers. Every situation is different.

Hope this helps!

Good Luck!

Lera25 [3.4K]2 years ago

3 0

You can subtract a negative number from a positive number or add a positive number to a negative number

Example:

4 - (7) = -3

-8 + 9 = 1

Whatever you are adding has to be greater than the number you are adding on too

Example:

-3 + 5 = 2

Notice how the five is greater than 3

Whatever you are subtracting has to be Greater than the number you are subtracting from.

Example:

5 - 8 = -3

Notice how the eight is greater than five.

The situation changed when adding or subtracting with negative numbers. Every situation is different.

Hope this helps!

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A fitness instructor tested the arm strength of members of a fitness dass to compare their scores to the national average. The t

tester [92]

Answer:

A) Median: 7

Mode: 8

Range: 5

B) Tyler is wrong, the greater the MAD value, the greater the variability

Step-by-step explanation:

A) Given the data set: 4,6,6,7,7,8,8,8,9

The median is 7 (the middle point in the ordered list)

The mode is 8 (the most repeated number)

The range is 5 (the difference between the highest and lowest values)

3 0

2 years ago

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal l

34kurt

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Temp. 174 176 177 178 178 179 180 181

Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74

Temp. 184 184 184 184 184 185 185 186

Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94

Temp. 186 186 186 188 188 189 190 192

Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16

Using the online linear regression calculator, the lie of best fit which models the data above is :

ŷ = 0.09386X - 15.55523

Where ;

X = independent variable

ŷ = predicted or dependent variable

- 15.55523 = intercept

0.09386 = gradient / slope

Point estimate when tank temperature is 186

ŷ = 0.09386(186) - 15.55523

ŷ = 17.45796 - 15.55523

ŷ = 1.90273

Residual error (y - ŷ), ŷ = 1.90273 when x = 186

(0.94 - 1.90273) = −0.96273

(1.85 - 1.90273) = −0.05273

(2.02 - 1.90273) = 0.11727

(2.64 - 1.90273) = 0.73727

To determine the proportion of observed variation in efficiency ratio, we find the Coefficient of determination R^2, which can be found using the online Coefficient of determination calculator : the r^2 value obtained is 0.4433.

5 0

2 years ago

I need help please this is due in about 5 minutes

strojnjashka [21]

5x^2= 2205

Divide both sides by 5
x^2=441

Find the square root of both sides
x= 21

4 0

3 years ago

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab

Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, $H_0$ : $\mu$ = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, $H_1$ : $\mu$ < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

s = sample standard deviation = 38.2 hours

n = sample size = 50

So, test statistics = $\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }$ ~ $t_4_9$

= -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0

3 years ago

|x - 1| > 4 is equivalent to which of the following?

Vsevolod [243]

Answer:

C. x > 5 or x < - 3

Step-by-step explanation:

|x - 1| > 4

Step 1:

x - 1 > 4

x > 4 + 1

x > 5

Step 2:

x - 1 < -4

x < -4 + 1

x < -3

HP: {x > 5 or x < -3} (c)

3 0

2 years ago