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3 years ago

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5.2x+8+2.1x-3 which is an equivalent expression by combining like terms A. 5.2x+8+2.1x-3 B. 5.2x+5+2.1x C 13.2x-0.9x D 7.3x+5

1 answer:

satela [25.4K]3 years ago

7 0

Answer: 7.3x + 5

Step-by-step explanation:

For us to solve the question, we have to solve 5.2x+8+2.1x-3 and the answer gotten will allow us know the equivalent expression.

= 5.2x+8+2.1x-3

= 5.2x + 2.1x + 8 - 3

= 7.3x + 5

Therefore, the correct answer is 7.3x + 5

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Cameron ran 3.8 miles per every 1.9 miles that Samantha ran and if Samantha and 7.6 miles how far did Cameron run

bija089 [108]

3.8 / 1.9 = x / 7.6....3,8 miles to 1.9 miles = x miles to 7.6 miles
cross multiply
(1.9)(x) = (3.8)(7.6)
1.9x = 28.88
x = 28.88 / 1.9
x = 15.2 <== Cameron ran 15.2 miles

8 0

3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$

Result is not significant.

3.

$(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378$

Result is not significant.

4.

$(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349$

Result is not significant.

5.

$(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773$

Result is not significant.

6.

$(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0$

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair die.

8 0

3 years ago

Becky spent 3/4 of an hour practicing the piano.She spent 1/2 of that time practicing her recital piece. What fractional part of

Ber [7]

Answer:

3/8

Step-by-step explanation:

First,

convert 3/4 of an hour to minutes;

If 1 hr = 60 mins

then 3/4 of 1hr = $\frac{3}{4} *60\\ \\ =45 mins.Next, If 1/2 of 3/4 was spent practicing recital piece, it means that it is half of 45 mins;1/2 *45 = 22.5 minsLastly, convert the 22.5 mins into a fraction of an hour;22.5/60 = [tex]\frac{3}{8}$

3 0

3 years ago

THIS IS EASY!!! (PLEASE HELP... ILL GIVE BRAINLIST)

rodikova [14]

Answer:

-7(9) < -56

Step-by-step explanation:

-7(9) is -63 which is less than -56 so anything above that is less than 56-

5 0

3 years ago

Read 2 more answers

A small school had 60% boys and the rest are girls if the number of boys is 48 what is the total number of boys and girls all to

Flura [38]

Hey there!

When dealing with percent word problems, the word "is" means equals. Let's see what we can do with your problem.

If the small school had 60% boys, it will have 40% girls. The number of boys <em>is</em> 48, which should equal 60% of the total, which we can model below, with t representing the total number of kids. In decimal form, 60% is 0.6.

0.6t=48

We divide both sides by 0.6.

t=80

Therefore, there are 80 total kids at the school.

We can also check our answer below.

BOYS: 0.6(80)= 48

GIRLS: 0.4(80)=32

48+32=80

I hope that this helps!

5 0

3 years ago