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kkurt [141]
2 years ago
14

The data represents the daily rainfall​ (in inches) for one month. Construct a frequency distribution beginning with a lower cla

ss limit of 00 and use a class width of .20. Does the frequency distribution appear to be roughly a normal​ distribution?
Mathematics
1 answer:
Lemur [1.5K]2 years ago
3 0
No no no no no no no no no no no no no no no no no
You might be interested in
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
Find the area of the figure to the nearest tenth.
Alisiya [41]

Answer:

36

Step-by-step explanation:

6 x 6 = 36

now to the nearest tenth it would be

36

so your answer is 36

5 0
3 years ago
Read 2 more answers
Henry is conducting an experiment with 800 bacteria. He found the exponential growth rate to be 2.7% per hour. What is the amoun
joja [24]
Let's formulate an equation first. For exponential growth, we follow the formula. b = 800, r - 0.027. Hence,

y = 800(1+0.027)^t

Then, at t = 21, y will be determined
y = 800 (1+0.027)^21
y = 1399.81 = 1400

Thus, the amount of bacteria present after 21 hours is 1400.
8 0
3 years ago
Find the slope of a line if the points (2,-1) and (5,3) are on the line
Romashka [77]

Answer:

\huge{ \boxed{ \sf{ \frac{4}{3} }}}

Step-by-step explanation:

\star{ \sf{ \: Let \: the \: points \: be \: A \: and \: B}}

\star{ \sf{ \: Let \: a(2,-1)  \: be \: (x1 \:, y1) \: and \: b (5,3)  \: be \: (x2 \:, y2)}}

\underline{ \sf{Finding \: the \: slope \: of \: the \: line}}

\boxed{ \sf{Slope =  \frac{y2 - y1}{x2 - x1} }}

\mapsto{ \sf{Slope =  \frac{3 - ( - 1)}{5 - 2} }}

\mapsto{ \sf{Slope =  \frac{3 + 1}{5 - 2}}}

\mapsto{ \sf{Slope =  \frac{4}{3} }}

Hope I helped!

Best regards! :D

~\text{TheAnimeGirl}

4 0
3 years ago
THIS IS URGENTE I AM BEING TIMED PLZ HELP, if you answer both correct i will mark you brainlyist
Lapatulllka [165]
Area would be 4.52 A= pi x radius squared
Circumference would be 7.54 C= 2 pi radius. Hope that makes sense.
4 0
3 years ago
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