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2 years ago

Please help I will give brainiest

Mathematics

2 answers:

Rom4ik [11]2 years ago

7 0

Answer:

1 in 9 chances of probability

Step-by-step explanation:

Ket [755]2 years ago

6 0

It is a 1 in 9 chances

hope this helps

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The cost for a hot dog and soft drink at a ballpark is $5.00. If the soft drink costs $2.50 less than the hot dog, find the cost

Elina [12.6K]

Answer:

Hot dog $3.75

Soft drink $1.25

Step-by-step explanation:

3.75 + 1.25 = 5.00

Difference

3.75 - 1.25 = 2.50

Hope this helps

8 0

2 years ago

Read 2 more answers

A nurse collected data about the average birth weight of babies in the hospital that month. Her data is shown using the dot plot

mixer [17]

Answer:

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X Frequency

8 0

8.1 0

8.2 2

8.3 2

8.4 2

8.5 3

8.6 0

8.7 1

8.8 0

8.9 1

9 0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

Minimum (shown at the bottom of the chart)
First Quartile (shown by the bottom line of the box)
Median (or the second quartile) (shown as a line in the center of the box)
Third Quartile (shown by the top line of the box)
Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

$[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3$ ,

Q₁ = 8.3.

The median value is,

$[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4$

Median = 8.4

The upper quartile is,

$[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5$ ,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

5 0

3 years ago

Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$