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2 years ago

Use the inequality symbol to compare 5+(-4)____________14+(-13)

Mathematics

2 answers:

Margarita [4]2 years ago

6 0

Both inequality is different positive and negative and it would be

NNADVOKAT [17]2 years ago

5 0

Answer:

5+(-4) ≤ 14+(-13)

Step-by-step explanation:

5+(-4) = 1

where 14 + (-13) = 1

seeing as 1 os whole we round down not up from whole numbers

Thus using the same sign reading left to right only as ≤ meaning less than or equal to for both sides.

OR write ≤ 1 ≥ as both inequalities = 1 and must be equal to or less than in same symbol reading left to right.

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I need help for this question

Nikitich [7]

V=V of cube- V pyramid= 8^3- 7^2*8/3=512-49*8/3=512-130,67=381.33 cm^3

7 0

2 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

2 years ago

What is the mode of this data on the quiz scores? 100, 98, 95, 92, 88, 85, 80, 75

larisa86 [58]

Answer:

There is no mode.

Step-by-step explanation:

Be sure to put in there is no mode, don't put in zero. If you put in zero, it will seem like ) is the mode, you know what I mean.

Be sure to mark me Brainliest!

7 0

2 years ago

Can Sombody help me real fast?

Stolb23 [73]

Answer:

45°

Step-by-step explanation:

Angles on a straight line add up to 180°.

135 + x = 180

x = 180 - 135

x = <u>45°</u>

3 0

2 years ago

Noah rides his bike for 14 miles in 1 hour. what is his average speed in miles per hour?

Delicious77 [7]

14 mph is the answer

5 0

2 years ago

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