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Ludmilka [50]
2 years ago
12

on his first 5 tests, che averaged 92 points. on his next 3 tests hw scored 94, 85, and 85 points. what was his average for all

8 tests?
Mathematics
1 answer:
Fofino [41]2 years ago
8 0

Answer:

90.5

Step-by-step explanation:

To get the average, we must add all the values and divide the sum by the amount of them:

  1. Add (92*5), 94, 85, 85 = 181
  2. Divide by the number of grades (8): 181/8

The answer will be 90.5, which is his average through all eight tests.

To do this in one step:

\frac{(92*5)+94+85+85}{8} = 90.5

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Given the sequence 4, 8, 16, 32, 64, ..., find the explicit formula.
Goshia [24]

Answer:

\sf 2^{(n+1)}

Step-by-step explanation:

Explicit formula is used to represent all the terms of the geometric sequence using a single formula.

  \sf \boxed{\bf t_n=ar^{(n-1)}}

Here, a is the first term.

r is the common ratio.

r = second term ÷ first term

4, 8,16,32,64,.....

    a = 4

    r = 8 ÷4 = 2

\sf t_n =4*2^{(n-1)}

    \sf = 4*2^n * 2^{(-1)}\\\\     = 4*2^n*\dfrac{1}{2}\\\\     = 2*2^n

   \sf = 2^{(n+1)}

4 0
1 year ago
A=1/2h(B+b); A=45, B=4, b=5
klemol [59]
Answer: h=10

Solving Steps:
45 = 1/2h x (4+5)
45 = 1/2h x 9
45 = 9/2h
10 = h
h = 10
4 0
3 years ago
What is the measure of
charle [14.2K]

Answer:

of what??

Step-by-step explanation:

?..............

4 0
3 years ago
Find the midpoint M of the two sets of points. (Use midpt. formula)
Setler79 [48]
3.5,1.75
7/2,3.5/2
Those 2
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2 years ago
The meat department of a local supermarket packages ground beef using meat trays of two sizes: 1 designed to hold 1 lb of meet a
r-ruslan [8.4K]

Answer:

c) 0.932

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

Step-by-step explanation:

Explanation:-

Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation  of 0.18 lbs.

size of the sample 'n' = 35

mean of the sample x⁻= 1.01lbs

standard deviation of the sample 'S' = 0.18lbs

<u>The 99% confidence intervals are given by</u>

(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )

The degrees of freedom γ=n-1 =35-1=34

tₐ =  2.0322

99% confidence interval for average weights of all packages sold in small meat trays

(1.01 - 2.0322 \frac{0.18}{\sqrt{35} } , 1.01+2.0322 \frac{0.18}{\sqrt{35} } )

( 1.01 - 0.06183 , 1.01+0.06183)

(0.932 ,1.071)

<u>Final answer</u>:-

<u>99% confidence interval for average weights of all packages sold in small meat trays.</u>

<u>(0.932 ,1.071)</u>

5 0
3 years ago
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