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Rama09 [41]
3 years ago
6

9y – 2 + y= 5y + 10​

Mathematics
1 answer:
RideAnS [48]3 years ago
8 0

Answer:

y=2.4

Step-by-step explanation:

9y-2+y=5y+10

(combine like terms)

10y-2=5y+10

(subtract 5y from both sides)

5y-2=10

(add 2 to both sides)

5y=12

(divide both sides by 5)

y=2.4

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What adds to get 3 but multiplies to get negative 18
Colt1911 [192]
6 and -3

6 + -3 = 3

6  x -3 = -18

Hope that helps! 
3 0
3 years ago
Simplify this equation<br> (3X^2+21X+33)/(X+5)
Amiraneli [1.4K]

Answer:

(3x^2+21x+33) / (x+5)

You can use long division:

(3x^2+21x+33)

--------------------

       (x+5)

=3x+6+ 3/x+5

Step-by-step explanation:

6 0
3 years ago
Which term is the perfect square of 3x^4
iragen [17]
The answer is 9x^8 
Hope it helped!!
7 0
3 years ago
Read 2 more answers
And also this question as well
gtnhenbr [62]

Answer:

It's the second one I believe, because it's not a straight line, and nearly all functions have curved lines

Step-by-step explanation:

4 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co
Ira Lisetskai [31]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

df = 103-1 = 102

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have T = 2.63.

Now, we need to find the standard deviation of the sample. That is:

s = \frac{0.56}{\sqrt{103}} = 0.055

Now, we multiply T and s

M = T*s = 2.63*0.055 = 0.145

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0
3 years ago
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