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3 years ago

Help will give brainlest

Mathematics

1 answer:

Tatiana [17]3 years ago

5 0

Answer:

The 2017 year is out of proportion.

Step-by-step explanation:

That graph should be at about half to 3/4 the height of the average heights of the other bars.

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PLEASE HELP I DONT UNDERSTAND

Annette [7]

Answer:

y = -x + 1

Step-by-step explanation:

eqn of line: y=mx+c

perpendicular, m1 x m2 = -1

m × 1 = -1

m = -1

y=-x+c

sub x=-5, y=6

6 = -(-5) +c

6 = 5 + c

c = 1

y = -x + 1

6 0

3 years ago

How many centiliters are in 6.02 decaliters

Vinil7 [7]

Well one decaliter is 1000 centiliters so 6.02 * 1000 = 6020 . so the answer is 6,020 centiliters .

8 0

3 years ago

Given f(x) = x + 3x + 2 and g(x) = x + 1, perform the indicated operations. In the

nadezda [96]

Answer: x² + 4x + 3

<u>Step-by-step explanation:</u>

f(x) = x² + 3x + 2

+ <u>g(x) = x + 1 </u>

(f+g)(x) = x² + 4x + 3

Note: This answer is based on f(x) having an x² (since the squared is missing from your question).

5 0

2 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago

HELP ASAP?!?!?!? QUADRATIC REGRESSION MODELS

CaHeK987 [17]

Answer:

4.87138

Step-by-step explanation:

i think

4 0

3 years ago