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3 years ago

A manatee population decreases by 15 manatees each year for 3 years. Find the total change in the manatee population.

Mathematics

2 answers:

agasfer [191]3 years ago

8 0

So 15 per year for 3 years or 3 times 15 or 45 total loss so the change is 45 difference or -45 in 3 years

Vladimir [108]3 years ago

6 0

$total\ change=3\ years\ * 15\ manatee\\\\
total\ change=45\\\\
In \ 3\ years\ time\ manatee\ population\ decreased\ by\ 45 manatees$

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diamong [38]

What problem there is no problem explain to me the problem

7 0

3 years ago

Christofle is making a doll house the doll house 6/10 meter high without the roof.Thelp roof is 15/100 meter high what will them

Alja [10]

You have to change the denominator of 6/10 into 100 then you will get 60/100 then you add 15/10+60/100

5 0

3 years ago

At a carnival, contestants are asked to keep rolling a pair of dice until they roll snake eyes. The number of rolls needed has a

Stells [14]

Answer:

The two numbers of rolls are 25.2 and 46.8.

Step-by-step explanation:

The Chebyshev's theorem states that, if X is a r.v. with mean µ and standard deviation σ, then for any positive number k, we have

$P (|X -\mu| < k\sigma) \geq (1-\frac{1}{k^{2}})$

Here

$(1-\frac{1}{k^{2}})=0.75\\\\\Rightarrow \frac{1}{k^{2}}=0.25\\\\\Rightarrow k=\sqrt{\frac{1}{0.25}}\\\\\Rightarrow k=2$

Then we know that,

$|X - \mu| \geq k\sigma,\\\\ \Rightarrow \mu - k\sigma \leq X \leq \mu + k\sigma$ .

Here it is given that mean (µ) = 36 and standard deviation (σ) = 5.4.

Compute the two values between which at least 75% of the contestants lie as follows:

$P(\mu - k\sigma \leq X \leq \mu + k\sigma)=0.75\\\\P(36 - 2\cdot\ 5.4 \leq X \leq 36 + 2\cdot\ 5.4)=0.75\\\\P(25.2\leq X\leq 46.8)=0.75$

Thus, the two numbers of rolls are 25.2 and 46.8.

8 0

3 years ago

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$

When $n\rightarrow \infty$ both $\displaystyle\frac{3}{n}$ and $\displaystyle\frac{1}{n^2}$ tend to zero and the upper sum tends to

$\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}$

8 0

3 years ago

The roof of a castle tower is shaped like a cone. The base of the cone is 24 m across and the height is 16 m. The slant height o

Elanso [62]

(a) Allow me to guide you on this one. Sketch a cone shape. At the circular base of the cone, sketch a straight line across through the middle which will be your diameter(most preferably where the slant height meet the circular base). Label it 24 m. From the apex of the cone, drop a straight line which is perpendicular to the diameter meeting it in the middle. Label that line 16 m. Label the line slanting from the top to the bottom x.

(b) The slant height of the roof is 20 m.
x^{2} =12^{2} + 16^{2} =144 + 256
x^{2} =400
x=√400
x=20m

6 0

3 years ago