All categories

3 years ago

Do you like math?? lets see

Mathematics

2 answers:

Archy [21]3 years ago

6 0

<h2>Helloooo!! Marie Here!!</h2>

The answer is here:

Usually, I hate math but sometimes depending on what mood I am or what grade level the math is, I like it!

If the math is easy then also yes, I do like it

If the math is hard then no, I do not like it

So I can/can't/won't/will like math at times

<h3>Hope This Helps! Have A GREATTT Day!!</h3>

Sloan [31]3 years ago

5 0

Answer:

YES BUT SOMETIMES NO

You might be interested in

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the le

Svet_ta [14]

Answer:

The maximum possible length, in centimeters, of the longest piece of rope is 134 cm

Step-by-step explanation:

Let the seven pieces be a,b,c,d,e,f,g

Average of 7 pieces of ropes =68

$Average = \frac{\text{Sum of all lengths}}{\text{No. of pieces}}$

$68= \frac{\text{Sum of all lengths}}{7}$

$68 \times 7 =\text{a+b+c+d+e+f+g}$

$476=\text{a+b+c+d+e+f+g}$ --A

We are given that the median length of a piece of rope is 84 centimeters.

We arrange the pieces in the ascending order

So, median will be the length of 4th piece

So, d = 84 cm

the length of piece a,b,c will be less than 84 and the value of e,f,g must be greater than or equal to 84

Let us suppose the length of a,b,c be x

Let us suppose the length of e,f be 84

The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope

So, g = 4x+14

Substitute the value in A

$476=x+x+x+84+84+84+4x+14$

$476=7x+266$

$\frac{476-266}{7}=x$

$30=x$

g = 4x+14=4(30)+14=134

Hence the maximum possible length, in centimeters, of the longest piece of rope is 134 cm

7 0

2 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

Find the interquartile range (IQR) of the data in the dot plot below.

kolbaska11 [484]

Answer:

<em>(IQR) interquartile range: 12 - 9 = 3</em>

Step-by-step explanation:

Median: 10

Lower quartile: 9

Upper quartile: 12

Interquartile Range: 12 - 9 = 3

5 0

3 years ago

What does it mean when someone says “ your nobody to lie to”?

timurjin [86]

Answer: You're not important basically

Step-by-step explanation:

4 0

2 years ago

Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m

suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week 1 2 3 4 5

Water Level (inches) − 1 1 4 1 3 8 2 1 2 -1 5 8 -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = $\frac{138 - 114}{2 - 1}$ = $\frac{24}{1}$ = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = $\frac{212 - 114}{3 - 1}$ = $\frac{98}{2}$ = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = $\frac{-158 - 114}{4 - 1}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = $\frac{-134 - 114}{5 - 1}$ = $-\frac{248}{4}$ = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = $\frac{212 - 138}{3 - 2}$ = $\frac{74}{1}$ = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = $\frac{-158 - 138}{4 - 2}$ = $-\frac{296}{2}$ = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = $\frac{-134 - 138}{5 - 2}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = $\frac{-158 - 212}{4 - 3}$ = $-\frac{370}{1}$ = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = $\frac{-134 - 212}{5 - 3}$ = $-\frac{346}{2}$ = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = $\frac{-134 + 158}{5 - 4}$ = $\frac{24}{1}$ = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0

2 years ago