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WINSTONCH [101]
2 years ago
12

Answer these please!! :( a) 5 + (-4) b) -9 + 16 c) 10 + (-14)

Mathematics
1 answer:
slamgirl [31]2 years ago
6 0

Answer:

a. 1

b. 7

c. -4

Step-by-step explanation:

a. 5 + (-4)

= 5 - 4

= 1

b. -9 + 16

= 7

c. 10 + (-14)

= 10 - 14

= - 4

If you further explanation how it come then please tell.

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alina1380 [7]

Answer:

A (50)

Step-by-step explanation:

Mean is the total numbers added up divided by the # of numbers.

There are 6 numbers.

51 + 60 + 80 + 32 + 47 + 30 = 300

300/6 = 50

Therefore, the answer is A.

7 0
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The formula for the area of a triangle is A=1/2bh. Solve the formula for h. A triangle has a base of 7 cm and an area of 28cm wh
faltersainse [42]
So, 7h/2=28.
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4 years ago
6) The height of any child in Mr. Diaz's kindergarten class is no greater than 46 inches.
kakasveta [241]

Answer:

full arrow going left from 46

Step-by-step explanation:

6 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

Compute the value of <em>n</em> as follows:

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

8 0
3 years ago
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