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2 years ago

An animal park has lions, tigers and zebras. 17% of the animals are zebras. What percentage are tigers? Show me your work!

Mathematics

1 answer:

antoniya [11.8K]2 years ago

3 0

Answer:

53% are tigers

Step-by-step explanation:

If you were referring to the question:

An animal park has lions, tigers and zebras. 17% of the animals are lions and 3/10 of the animals are zebras. What percentage are tigers? Show me your work!

We can then solve it by considering what is 3/10 in percent form.

When you think of percent, just remember that it is a portion for every 100. So think of a fraction that is proportional to 3/10 that is a fraction of 100:

$\dfrac{3}{10} = \dfrac{x}{100}\\\\\dfrac{300}{10}=x\\\\30 = x$

So 3/10 = 30/100 which in percentage form is 30%.

Now if:

17% = lions

30% = zebras

We can get the percentage by looking for how many out of 100% is left.

100% - (17% + 30%)

100% - 47%

53%

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ABC is a triangle and PQ is parallel to BC. BC=12.6cm, PQ=8.4 cm and AQ= 7.2 cm. Find AC

nlexa [21]

With a line parallel to a side through a triangle we get similar triangles so a proportion.

AC : BC = AQ : PQ

AC = BC AQ / PQ

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2 years ago

Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .

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Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

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2 years ago

A right pyramid with a square base has a base edge length of 12 meters and slant height of 6StartRoot 2 EndRoot meters. The apot

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Answer:

Volume of the right pyramid = 288 m²

Step-by-step explanation:

Volume of the pyramid = $\frac{1}{3}\times {\text{Area of the rectangular base}\times height(h)$

From the ΔAOB,

By Pythagoras theorem,

AB² = AO² + OB²

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Now volume of the pyramid = $\frac{1}{3}[(Length)(width)]\times height$

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= 288 m²

Therefore, volume of the right pyramid is 288 m².

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2 years ago

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Line q will be graphed on the same grid. The only solution to the system of linear equations formed by lines n and q occurs when

bija089 [108]

Answer:

$f(x) = k(x - \frac{3}{2})$

Step-by-step explanation:

Line q will be graphed on the same grid. The only solution to the system of linear equations formed by lines n and q occurs when x = $\frac{3}{2}$ and y = 0.

Now, as x = $\frac{3}{2}$ is a solution of the equation, y = f(x) = 0, so, $(x - \frac{3}{2})$ will be a factor of the linear function y = f(x).

Therefore, we can write the possible equation for the line q will be

$f(x) = k(x - \frac{3}{2})$ . (Where k is any constant} (Answer)

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2 years ago

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$

$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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2 years ago