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Paraphin [41]
2 years ago
14

Which scenario fits the compound inequality?

Mathematics
2 answers:
lisov135 [29]2 years ago
8 0

Answer: In terms of percent, the salinity of most sea water is between 2.5 and 4.5

Step-by-step explanation:

krok68 [10]2 years ago
6 0

Answer:

In terms of percent, the salinity of most sea water is between 2.5 and 4.5.

Step-by-step explanation:

A compound inequality has at least two inequalities where they are separated by either and or "or".

As per definition, the correct answer is - In terms of percent, the salinity of most sea water is between 2.5 and 4.5.

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Which table of ordered pairs represents a proportional relationship?​
34kurt

Answer:

its c

Step-by-step explanation:

5 0
2 years ago
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Hadley paddled a canoe 2/3 mile in 1/4 hour. How fast did Hadley paddle in miles per hour ?
Vitek1552 [10]

Answer:2 2/3 miles per hour

Step-by-step explanation:to get one mile you multiply 1/4 by 4 and to keep 1/4 and 2/3 proportional you also have to multiply 2/3 by 4

5 0
2 years ago
A pharmacy uses 4×10−3 liter of an active ingredient in one dose of a medication. The active ingredient comes in a 2-liter bottl
Firlakuza [10]
Since each bottle is made up of 2-liter of the active solution, in order to determine the number of doses of the medication that can be made, we just have to divide 2 L by 4x10^-3.
                  number of doses = (2 L) / (4x10^-3L/dose) = 500 doses
Therefore, the pharmacy can make 500 doses of the medication. 
3 0
2 years ago
6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability th
Maru [420]

Answer:

(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.

(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

Step-by-step explanation:

Let <em>X</em> = number of heads.

The probability that a head occurs in a toss of a coin is, <em>p</em> = 0.50.

The coin was tossed <em>n</em> = 100 times.

A random toss's result is independent of the other tosses.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and <em>p</em> = 0.50.

But the sample selected is too large and the probability of success is 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of \hat p<em> </em>(sample proportion of <em>X</em>) if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=100\times 0.50=50>10\\n(1-p)=100\times (1-0.50)=50>10

Thus, a Normal approximation to binomial can be applied.

So,  \hat p\sim N(p,\ \frac{p(1-p)}{n})

\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05

(a)

Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:

P(0.30

                              =P(-4

Thus, the probability that proportion of heads is between 0.30 and 0.70 is 1.

(b)

Compute the probability that proportion of heads is between 0.40 and 0.65 as follows:

P(0.40

                              =P(-2

Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

6 0
2 years ago
Find the sum of the first 30 terms of the sequence below an=3n+2
Solnce55 [7]
First off, let's find the 1st term's value, and the 30th term's value,

\bf a1=3(1)+2\implies a1=5\qquad \qquad \quad   a30=3(30)+2\implies a30=92\\\\&#10;-------------------------------\\\\&#10;~~~~~~~\textit{ Sum of an arithmetic sequence}\\\\&#10;S_n=\cfrac{n(a1+an)}{2}~ &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a1=\textit{first term's value}\\&#10;----------\\&#10;a1=5\\&#10;a30=92\\&#10;n=30&#10;\end{cases} \implies S_{30}=\cfrac{30(5+92)}{2}&#10;\\\\\\&#10;S_{30}=15(97)
6 0
2 years ago
Read 2 more answers
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