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olya-2409 [2.1K]
3 years ago
12

10. Tell whether the sequence is arithmetic. If it is, what is the common difference? (1 point)

Mathematics
1 answer:
Stella [2.4K]3 years ago
6 0

Answer:

No

Step-by-step explanation:

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Plz help need answer for angle y
nalin [4]
I got the answer to be 42.
3 0
3 years ago
Pls help!! I need the answer immediately
Strike441 [17]

Answer:

i). x³ + 9x² + yz - 15

ii). -21m³np - 8p⁵q + mnp + 4mn + 100

Step-by-step explanation:

Question (38)

i). Two expressions are -5x² - 4yz + 15 and x³+ 4x²- 3yz

  By subtracting expression (1) from expression (2) we can the expression by addition which we can get expression (1).

 (x³+ 4x²- 3yz) - (-5x² - 4yz + 15) = x³ + 4x² - 3yz + 5x² + 4yz - 15

                                                    = x³ + 9x² + yz - 15

ii). -15m³np + 2p⁵q - 6m³pn + mnp + 4mn - 10qp⁵+ 100

  = (-15m³np - 6m³np) + (2p⁵q - 10qp⁵) + mnp + 4mn + 100

  = -21m³np - 8p⁵q + mnp + 4mn + 100

4 0
3 years ago
Plzz helpp guys
Kitty [74]

Answer:

y = 1500(0.024)^t

Step-by-step explanation:

Given

a = 1500km -- initial area

b = 2.4\% -- rate

t = time --- (years)

Required

Represent as an exponential function

An exponential function has the form:

y = ab^t

Where

a = initial\ value

b = rate and b \ne 1

t= time

Substitute values for a and b, we have:

y = 1500 * (2.4\%)^t

Convert percentage to decimal

y = 1500 * (0.024)^t

y = 1500(0.024)^t

Hence, the exponential function that represents the scenario is: y = 1500(0.024)^t

6 0
3 years ago
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ab
zhenek [66]

Answer:

test statistic is ≈ -0.36

p-value is  ≈ 0.64

There is no significant evidence that the average golfer can hit the ball more than 235 yards on average.

Step-by-step explanation:

a hypothesis test where H_0: mu = 235 and H_1:mu > 235

test statistic can be calculated as follows:

z=\frac{X-M}{\frac{s}{\sqrt{N} } } where

  • sample mean driving distance (233.8 yards)
  • M is the average expected distance that the average golfer can hit the ball under null hypothesis. (235 yards)
  • s is the standard deviation (46.6 yards)
  • N is the sample size (192)

Then test statistic is z=\frac{233.8-235}{\frac{46.6}{\sqrt{192} } } =-0.3568

p-value is  0.64 >0.05

There is no significant evidence that the average golfer can hit the ball more than 235 yards on average.

6 0
3 years ago
What is the same as 1 cm to 1 km
lana [24]
Pretty easy little one hehe, hoped this helped 1m=1000mm
6 0
3 years ago
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