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Fittoniya [83]
4 years ago
14

What are the terms in the expression 5a+3b+1 PLZ HELPPPPPPPPPPPP

Mathematics
2 answers:
slega [8]4 years ago
7 0
5a + 3b + 1
There are 3 terms in this expression...they are (5a) , (3b), and 1
juin [17]4 years ago
3 0
1 is a term and 5a and 3b are terms because 1 is a single number and 5a and 3b both have variables 

:)
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12 - 8 x 9 +16<br> Order of operation
olga nikolaevna [1]
Answer:

12-72+16
12-88
-76
7 0
4 years ago
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
What’s the correct answer for this?
charle [14.2K]

Answer:

YT AND RQ

Step-by-step explanation:

7 0
3 years ago
Without using calculator find :<br> cos(225) sin(315) + sin( –270) tan(405)
Ronch [10]

Answer:

1.5

Step-by-step explanation:

225° bisects Q III

315° bisects Q IV

-270° = 90°

405° = 45° bisects Q I

cos(225) sin(315) + sin( –270) tan(405) = ?

      (-½√2)(-½√2) + 1(1) = ?

                          0.5 + 1 = 1.5

6 0
3 years ago
Read 2 more answers
What is the opposite of the number -2.5? How does the position of -2.5 compare to its opposite on the number line?
charle [14.2K]

Answer:

-2.5 and 2.5 compare to eachother on the number line because they are the exact same distance away from 0.

Step-by-step explanation:

0 is in the very middle and to get -2.5 you go to the left 2.5 paces but to get to +2.5 you go to the right 2.5 paces. Either way you have to move Exactly 2.5 paces either to the left or right depending on if you want a positive or negative.

8 0
3 years ago
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