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3 years ago

PLEASE HELP ME?!

Mathematics

2 answers:

Stells [14]3 years ago

7 0

It's 44 square units as you can draw lines down from the corners on the top to make two triangles. then cut one off, flight it and add it too the other side. count along 11 and times it by 4 up.

slavikrds [6]3 years ago

4 0

87 square units hope that this helped

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????????????????????????

choli [55]

Answer: 12

Step-by-step explanation:

$\frac{6}{18}=\frac{4}{c}$

Solve for c

$c=\frac{4*18}{6}\\ c=\frac{72}{6} \\c=12$

3 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

you plant 15 tulip bulbs and 10 of them grow into flowers to bloom. If you plant 21 more re tulips bulbs, predict how many of th

harina [27]

Answer:

14 more flower would bloom if 21 more tulip bulb is planted.

Step-by-step explanation:

Given ratio for tulip bulbs and flowers is 15:10.

Now, finding number of flower to bloom if 21 tulip bulb are planted.

Assume if 21 tulip bulbs are planted then number of flower would bloom be"x"

ratio given: $\frac{Tulip\ bulbs}{ Flower} = \frac{15}{10} = \frac{21}{x}$

Cross multiplying the ratio to get:

⇒x= $\frac{21\times 10}{15} = \frac{210}{15}$

∴ $x= 14\ flowers$

Hence, 14 flowers would bloom if 21 tulip bulb are planted.

If total tulip bulb is $15+21= 36$ , then $10+14= 24\ flowers$ would bloom.

5 0

3 years ago

Suppose Skyler invests in an annuity that pays 3.2% annual interest, compounded monthly. If she contributes $155 every month for

Gennadij [26K]

Answer:

$3286.47

Step-by-step explanation:

0.032/12

P_10 = 21886.46638739

155 * 12 * 10 = 18600

21886.46638739 = 18600 + I_10

I_10 = 3286.46638739

round answer to nearest cent

$3286.47

8 0

2 years ago

I am having trouble with this could somebody give me a hand

MrRa [10]

check the picture below.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+100}t\stackrel{\stackrel{c}{\downarrow }}{+12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left( -\cfrac{100}{2(-16)}~~,~~12-\cfrac{100^2}{4(-16)} \right)\implies \left( \cfrac{25}{8}~~,~~12+\cfrac{625}{4}\right) \\\\\\ \left(\cfrac{25}{8}~,~\cfrac{673}{4} \right)\implies \left(\stackrel{\stackrel{\textit{took this long}}{\downarrow }}{3\frac{1}{8}}~~,~~\stackrel{\stackrel{\textit{went this high}}{\downarrow }}{168\frac{1}{4}} \right)$

5 0

3 years ago