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2 years ago

What is the approximate percent decrease when a fish count goes down from 850 to 500?

2 answers:

8 0

To get this, you divide the smaller number by the larger one and get 0.5882... 500 is about 59% of 850, so to find the percent lost, you subtract that from 100 to get 41% decrease

Oksana_A [137]2 years ago

6 0

<span>What is the approximate percent decrease when a fish count goes down from 850 to 500?</span>The answer is 40%

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Mr. and Mrs. Smith's children all play different sports. They want to buy a combination of soccer training sets,which each cost

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1 year ago

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Joanna pays $40 plus a $2 surcharge each month for her high-speed Internet service. Which table BEST represents the relationship

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2 years ago

Which equation is the inverse of y = 16x2 + 1? y = plus-or-minus StartRoot StartFraction x Over 16 EndFraction minus 1 EndRoot y

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Answer:

The inverse is ±sqrt((x-1))/ 4

Step-by-step explanation:

y = 16x^2 + 1

To find the inverse, exchange x and y

x = 16 y^2 +1

Then solve for y

Subtract 1

x-1 = 16 y^2

Divide by 16

(x-1)/16 = y^2

Take the square root of each side

±sqrt((x-1)/16) = sqrt(y^2)

±sqrt((x-1))/ sqrt(16) = y

±sqrt((x-1))/ 4 = y

The inverse is ±sqrt((x-1))/ 4

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1 year ago

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

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Answer:

$\frac{2}{9}$

Step-by-step explanation:

Given :

Juan rolled a six-sided number cube 18 times.

The number two occurred four times.

To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?

Solution:

Total events = number of times cube rolled = 18

Favorable events = The number two occurred four times. = 4

So, Experimental probability = $\frac{\text{Favorable events}}{\text{Total events}}$

= $\frac{4}{18}$

= $\frac{2}{9}$

Thus the experimental probability of rolling a two was $\frac{2}{9}$

So, Juan’s experimental probability was incorrect.

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2 years ago

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In a normally distributed data set a mean of 55 where 99.7% of the data fall between 47.5 and 62.5, what would be the standard d

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Given that normally distributed data set has a mean of 55 and 99.7% of data fall between 47.5 and 62.5.

Let s be the standard deviation of data set.

Since 99.7% data fall within 3 standard deviations of mean, z-value for 47.5 and 62.5 has an absolute value of 3.

That is |z|=3

But z= $\frac{x-mean}{standard deviation}$

Let us plugin x=47.5 and mean =55 and equate it to 3.

That is $|\frac{47.5-55}{s}| = 3$

$|\frac{-7.5}{s} | =3$

Since x is always positive ( being standard deviation), $|\frac{-7.5}{s} | = \frac{|-7.5|}{s} = \frac{7.5}{s}$

Hence $\frac{7.5}{s}= 3$

$s=\frac{7.5}{3} = 2.5$

We will get same value with 62.5 as well.

Hence standard deviation of data set is 2.5.

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2 years ago