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scoundrel [369]
3 years ago
7

The line plot shows the heights of plants in an experiment, Which statement is supported by the data in the line plot? More plan

ts aro 56 inches tall than aro 62 Inches tall. Plant Experiment There are 25 plants in the experiment. xXx XX XXXX The range of heights is 63 inches. Most of the plants are under 61 inches tall.​
Mathematics
1 answer:
Advocard [28]3 years ago
7 0

Answer:Answer the range of heights is 63 inches

Step-by-step explanation. Better do ur I ready test stop cheating

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What is the answer?please help
White raven [17]

Answer:

d

Step-by-step explanation:

4 0
3 years ago
Pairs that satisfy function y=2x+1
evablogger [386]

Test a pair from each table by substituting their values into the given equation and solving.

A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1

B) y = 2x + 1 .......... 1 = 2(0) + 1 .......... 1 = 1

C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1

D) y = 2x + 1 .......... -2 = 2(0) + 1 .......... -2 ≠ 1


The only pair that satisfied the equation was from answer choice B. Therefore, B is the correct answer.

8 0
3 years ago
What is the measure of angle c In degrees?
Irina-Kira [14]
C = 64
73 + 43 = 116
180 - 116 = 64

4 0
3 years ago
Read 2 more answers
19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical
Tems11 [23]

EXPLANATION

Since we have the function:

f(x)=\frac{x^2-8x+15}{x^2}

Vertical asymptotes:

For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.

Taking the denominator and comparing to zero:

x+5=0

The following points are undefined:

x=-5

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1y=\frac{1}{1}\mathrm{The\:horizontal\:asymptote\:is:}y=1

In conclusion:

\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1

4 0
1 year ago
How do you work out a percentage
Anni [7]
You can find percentages with a calculator, or without a calculator:

Non-calculator:
To find a percentage of a number without a calculator, you divide the number by the percentage's number equivalent. For example, to find 10% of a number you divide by 10, to find 25% of a number you divide by 4 (because 25% is 1/4 of 100%), to find 50% you divide by 2, and 1% is dividing by 100, etcetera. You can then use the initial percentage's to find more complex percentages, so to find 30% you would divide by 10 and then multiply that by 3, and 75% would be dividing by 4 and multiplying that be 3.

Calculator:
If you have a calculator when you are asked to find the percentage, there is a much simpler way. All percentages have a decimal equivalent because you can picture all percentages as being fractions over 100, and therefore if you divide that number by 100 you get its decimal. On a calculator, if you multiply a number by a decimal multiplier, you get that percentage. For example, if you want to find 25% of a number on a calculator, you would multiply that number by 0.25.

I hope this explains things well enough! Let me know if I missed anything out :)
3 0
3 years ago
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