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mixer [17]
3 years ago
5

If a relationship can be written as s=−5+2t, how does s change if t is increasing

Mathematics
2 answers:
Studentka2010 [4]3 years ago
8 0

Answer:

s increases

Step-by-step explanation:

2t is a positive value so it will make the value of a increase

Alenkinab [10]3 years ago
3 0
S would also increase if t is being increased
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Part D<br> What is the median number of words in Darcy's essays?
NARA [144]

Answer:

210.8 (mean for Luis), 241.5 (mean for Darcy)

Step-by-step explanation:

Luis 178 + 213 + 198 + 245 + 236 + 198 + 221 + 253 + 189 + 177 = 2108

2108/10 = 210.8 (mean for Luis)

Darcy 231 + 210 + 245 + 259 + 286 + 245 + 231 + 244 + 236 + 228 = 2415

2415/10 = 241.5 (mean for Darcy)

4 0
2 years ago
What is the median of this data set?
lora16 [44]

Answer:

Step-by-step explanation:

7 0
3 years ago
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15 points !!! will mark brainiest if you get correct !!!
lozanna [386]

Step-by-step explanation:

1. C:arithmetic, 34, 39, 44

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3. C: geometric, 1, 1/3, 1/9

4 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
A shelter has enough food to feed an average of 80 hungry people a day for 14 days. If an average of 32 people are at the shelte
egoroff_w [7]

Multiply 80 by 14 to get how many total people they can feed.

80 times 14 is 1120.

Now, divide 1120 by 32 to get how many days the food will last if 32 people come every day.

1120/32 = 35.

The answer is B, 35 days

4 0
3 years ago
Read 2 more answers
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