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faltersainse [42]
2 years ago
10

Please help me thank you so much!

Mathematics
2 answers:
ad-work [718]2 years ago
5 0

Answer:

610 squares on the final day.

Step-by-step explanation:

490 Sunday
510 Monday
530 Tuesday
550 Wednesday
570 Thursday
590 Friday
610 Saturday

Andreas93 [3]2 years ago
4 0
Should have ate zero if I’m not wrong
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Solve the following equation.<br><br> 2/3x + 6 = 1/2x + 1/4x<br><br> 2<br> 6<br> 14 2/5<br> 72
Vladimir79 [104]
Answer: 72
heyy, umm basically group like terms together. E.g. 6= 1/2x + 1/4x - 2/3x. Then, just solve for x e.g. 1/2 + 1/4 - 2/3 and use the answer to equal it to 6. Which should be 1/12x=6... pretty sure haha. Then divide. E.g.6/1/12 for x. In this case I got 72.. :)
5 0
3 years ago
Y= 2x + 7 y= -4x - 5 What is the solution for the system of equations?
Oksi-84 [34.3K]

Answer:

x = -2, y = 3.

as an ordered pair it is (-2, 3).

Step-by-step explanation:

y= 2x + 7

y= -4x - 5

As both right hand expression are = y we have:

2x + 7 = -4x - 5

6x = -12

x = -2

Substitute for x in equation 1:

y = 2(-2) + 7 = 3.

7 0
3 years ago
Read 2 more answers
2/5 in Fraction form?
hram777 [196]

Answer:

2/5 is fraction form.

0.4 is decimal form btw.

5 0
3 years ago
Combine like terms.
Vesna [10]
To combine these terms you would add up the a's and b's.
4a+b+a is the same as 4a+a+b. Anything without a constant in front has a value of 1. You have 4a's and you need to add another a. That equals 5a's. The b is left alone since you can only add like variables.
The answer would be C) 5a +b
5 0
3 years ago
Read 2 more answers
A student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly se
Darina [25.2K]

Answer: (17.42, 20.78)

Step-by-step explanation:

As per given , we have

Sample size : n= 9

\overline{x}=19.1 years

Population standard deviation is not given , so it follows t-distribution.

Sample standard deviation : s= 1.5 years

Confidence level : 99% or 0.99

Significance level : \alpha= 1-0.99=0.01

Degree of freedom : df= 8  (∵df =n-1)

Critical value : t_c=t_{(\alpha/2,\ df)}=t_{0.005,\ 8}= 3.355

The 99% confidence interval for the population mean would be :-

\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}

19.1\pm (3.355)\dfrac{1.5}{\sqrt{ 9}}\\\\=19.1\pm1.6775\\\\=(19.1-1.6775,\ 19.1+1.6775)\\\\=(17.4225,\ 20.7775)\approx(17.42,\ 20.78)

Hence, the 99% confidence interval for the population mean is (17.42, 20.78) .

7 0
3 years ago
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