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3 years ago

15

94% of salmon pass through a single dam unharmed. By what percent does the number of salmon decrease when passing through a sing

le dam?

2 answers:

professor190 [17]3 years ago

7 0

Answer:

6%

Step-by-step explanation:

Vanyuwa [196]3 years ago

3 0

Answer:

They decrease by 6%

Step-by-step explanation:

100%-94%=6%

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The formula for the volume of a sphere is v= 4/3 3.14rWhat is the formula solved for r?

prohojiy [21]

Answer:

$r = \frac{3}{4\pi}V$

Step-by-step explanation:

First, the formula for the volume of a sphere is:

$V = \frac{4}{3}\pi r^3$

$\frac{3}{4}V = \pi r^3$

$\frac{3}{4}V\frac{1}{\pi} = r^3$

$r^3 = \frac{3}{4\pi}V$

$r = \sqrt[\leftroot{-2}\uproot{2}3]{\frac{3}{4\pi}V}$

<u><em>If there is any steps you are unsure of, feel free to ask in the comments.</em></u>

6 0

3 years ago

Quadrilateral ABCD is inscribed in a circle with angle measures m∠A = (11x − 8)°, m∠B = (3x2 + 1)°, m∠C = (15x + 32)°, and m∠D =

dimulka [17.4K]

Answer:

No for all of the 4 values

Step-by-step explanation:

The sum of the internal angles of a quadrilateral is always 360°, so we have that:

m∠A + m∠B + m∠C + m∠D = 360

11x - 8 + 3x2 + 1 + 15x + 32 + 2x2 - 1 = 360

5x2 + 26x - 320 = 0

Solving the quadratic function using Bhaskara's formula, we have:

Delta = 26^2 + 4*5*320 = 7076

sqrt(Delta) = 84.12

x1 = (-26 + 84.12)/10 = 5.81

x2 = (-26 - 84.12)/10 = -11.01 (this value will generate negative angles, so it is not valid).

Now, finding the angles, we have:

m∠A = 11*5.81 - 8 = 55.91°

m∠B = 3*(5.81)^2 + 1 = 102.27°

m∠C = 15*5.81 + 32 = 119.15°

m∠D = 2*(5.81)^2 - 1 = 66.51°

As all these values are different from the values shown, the answer is No for all of them.

8 0

3 years ago

-3x+5y=-9 and 4x+8y=12

insens350 [35]

-3x + 5y = -9

4x + 8y = 12

Simplify the second,

x + 2y = 3

Times three,

3x + 6y = 9

Add to the first equation,

11y = 0

y = 0

-3x + 5(0) = -9

x = 3

Answer: x=3, y=0

4 0

3 years ago

Read 2 more answers

Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

$\hat p_1 =0.768$ represent the estimated proportion for 1

$n_1=92$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.646$ represent the estimated proportion for 2

$n_2=95$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

7 0

3 years ago

As the teams fundraiser,Agnes and Betty both sold candy bars at the end of the fundraiser Agnes determined that she sold 8 more

Romashka-Z-Leto [24]

Answer:

356

Step-by-step explanation:

5 0

2 years ago