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maw [93]
3 years ago
12

Please help me! I need part A and part B. Please help me! PLEASE!!

Mathematics
2 answers:
likoan [24]3 years ago
7 0
The answer is 8654 I did the math
ehidna [41]3 years ago
4 0

Answer:

JUST TO START I COMMENT ON THIS SORRY

Step-by-step explanation:

4366+766

5251551=8654

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Divide using long division.<br> (2x2 + 7x + 6) : (x + 2)
Usimov [2.4K]

Answer:

2x+3

Step-by-step explanation:

-2x2 - 7x - 6

--------------------

-x-2

-2x2 - 7x - 6  =   -1 • (2x2 + 7x + 6)

 -x - 2  =   -1 • (x + 2)

 2x2 + 7x + 6

x • (2x+3) and      (x+2)  •  (2x+3)

answer- 2x+3

4 0
3 years ago
Read 2 more answers
On a number line 8.63 would be located where
Oksanka [162]

Answer:

It would be slightly higher than 8.5 and lower than 9.

Step-by-step explanation:

8.6 is bigger than 8.5, but it's smaller than 9, so you would place it in between the two but closer to the 8.5 mark.

4 0
3 years ago
A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The
TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 16 ​salmon, and to discount outlet stores in pallets of 64 salmon. To forecast​ costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts​ (a) and​ (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: \frac{s.d}{\sqrt{u}}

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{9}}

= \frac{3}{3} = 1

The standard deviation = 1

ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{16}}

= \frac{3}{4} = 0.75

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{64}}

= \frac{3}{8} = 0.375

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

5 0
3 years ago
the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i
nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

A = \frac{1}{2}\cdot b\cdot h (Eq. 1)

Where:

A - Area of the triangle, measured in square centimeters.

b - Base of the triangle, measured in centimeters.

h - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt} (Eq. 2)

Where:

\frac{dA}{dt} - Rate of change of area in time, measured in square centimeters per minute.

\frac{db}{dt} - Rate of change of base in time, measured in centimeters per minute.

\frac{dh}{dt} - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

\frac{1}{2}\cdot\frac{db}{dt}\cdot h =  \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

b = \frac{2\cdot A}{h}

If we know that A = 130\,cm^{2}, h = 15\,cm, \frac{dh}{dt} = 2.5\,\frac{cm}{min} and \frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}, the rate of change of the base of the triangle in time is:

b = \frac{2\cdot (130\,cm^{2})}{15\,cm}

b = 17.333\,cm

\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)

\frac{db}{dt} = -2.262\,\frac{cm}{min}

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0
2 years ago
A 52 card deck of cards is dealt from the top of the deck. what's the probability that it is a red?
Semmy [17]
The probability it is red is 1/2 because there are 2 suits of both red and black making each color have 26 cards so equal amounts equals 1/2
6 0
3 years ago
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