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victus00 [196]
3 years ago
15

The wankel T. rex is in display at the Smithsonian Washington DC scientists estimate that it was 18 years old when it died. Abou

t how much did it weight?

Mathematics
1 answer:
Phoenix [80]3 years ago
5 0

Answer:

About 5,000 kilograms

Step-by-step explanation:

Look on the x-axis for 18 years. Then go up to where the plotted line is on 18 years to find the y-axis or the mass.

You might be interested in
The Wall Street Journal reported that the median salary for middle-level manager jobs was approximately $85,000 (The Wall Street
vredina [299]

Answer:

Median = $80,000 ; Mean = $84,000 ; Q1 = 67,000 ; Q3 = 106,000

Step-by-step explanation:

Given the data :

108 83 106 73 53 85 80 63 67 75 124 55 93 118 77

Reorderd data :

53, 55, 63, 67, 73, 75, 77, 80, 83, 85, 93, 106, 108, 118, 124

The median salary:

1/2(n+1)th term

Sample size, n = 15

1/2 (15 + 1)th term

1/2(16)th term

8th term = 80

Median is $80,000

Median reported by Wall Street Journal approximately $85000

The obtained and reported values are close

The mean annual salary :

ΣX / n = 1260 / 15 = 84

Mean annual salary = $84,000

Mean is the average of all earnings combined, median is the mid value of he data.

First and third quartile :

Q1 = 1/4(n+1)th term

Q1 = 1/4(16)th term

Q1 = 4th term = 67

Third quartile:.

Q3 = 3/4(n+1)th term

Q3 = 3/4(16)th term

Q3 = 12th term = 106

3 0
3 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
Help me plzzzz. it's confusion
stich3 [128]
6) \sqrt[4]{16} = 2;
3 0
4 years ago
WHATS THE AREA OF THE TRAPIZOID?!
Helen [10]

Answer: 45.. i think.

Step-by-step explanation: I couldn't really get the measurements from this photo so here is what I got. I used the formula A=a+b over 2 times h. I used 7=a, b=11,h=5. I added 7 and 11 to get 18, then divide that by 2, which equals 9. Then 9 times 5 = 45. I hope this somewhat helps.

3 0
3 years ago
Need help on thisss (:
Setler79 [48]

The area is 77 cm

Area= ab/2

area and base=14·11 /2

14 x 11 / 2 =77

4 0
3 years ago
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