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3 years ago

5

Isla was researching kiwi birds. She categorized each bird she observed by species and gender. The two-way frequency table below

shows data from observations of 200 kiwis.
Which of the following statements is true about the kiwis observed?
(Choice A)
There were the most observations of female tokoeka kiwis.

(Choice B)
Great spotted kiwis were more likely to be male than female.

(Choice C)
More than half of the kiwis were Northland brown kiwis.

(Choice D)
Northland brown kiwis were more likely to be female than tokoeka kiwis were.

2 answers:

Ludmilka [50]2 years ago

8 0

Answer:

A.

I hope it helps

CaHeK987 [17]2 years ago

3 0

A is the answer my dude

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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

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The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

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2 years ago

5(-42/-7) + (-15) - (-3)

siniylev [52]

Answer:

2058

Step-by-step explanation:

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2 years ago

(Do not use spaces. Use to represent exponents. Example 2^3 is 22.)

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Answer: y=6^x-3

It is a exponent form of graph, so first:

y=a^x-b

When b=0, the asymptote is y=0 but as the asymptote given is y=-3, b=-3

Second:

the y value increases 6, when x changes 0 to 1, so a=6

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2 years ago

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Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit

coldgirl [10]

Changing the equation into slope form: $y = mx + c$ , where $m$ is the slope [gradient] and $c$ is the y-intercept.

$2x+3y=1470$
$3y = -2x+1470$
$y= - \frac{2}{3}x+ \frac{1470}{3}$
$y=- \frac{2}{3}x+490$

The gradient is $- \frac{2}{3}$ and y-intercept is at $y=490$

Graphing $y=- \frac{2}{3}x+490$ using slope-intercept method:
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
   $0 = - \frac{2}{3}x+490$
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   $x = \frac{1470}{2}$
   $x = 735$
So, x-intercept is at (735, 0)

The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735

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

Next month's profit equation

$2x+3y=1593$

Rewriting this into slope-equation form

$3y = -2x+1593$
$y=- \frac{2}{3}+ \frac{1593}{3}$
$y= - \frac{2}{3}+531$

The gradient, $m$ , equals to $- \frac{2}{3}$
The y-intercept, $c$ , equals to 531

The equation still has the same gradient with last month's profit equation but different y-intercept.

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A linear graph show points of (0, 300) and (450, 0)

We work out the slope:
$\frac{300-0}{0-450} = \frac{300}{-450}=- \frac{20}{30} =- \frac{2}{3}$

Y-intercept at x = 0, so it's at y = 300

Equation $y = - \frac{2}{3}+300$

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3 years ago

Find the number the comes next 1.2.6,21.88,?

Nataly_w [17]

Answer

455

Step-by-step explanation:

(1×1) + 1 = 2

(2×2) + 2 = 6

(6×3) + 3 = 21

(21×4)+4 = 88

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