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3 years ago

8

Mount Rushmore is a sculpture that was carved using a model with a scale of 1 in :1 ft. If the model of George Washington’s face

was 5 ft tall, how tall is his face on Mount Rushmore?

2 answers:

nordsb [41]3 years ago

3 0

Answer:

60 feet but i think the question is written wrong

Step-by-step explanation:

SIZIF [17.4K]3 years ago

3 0

134ft tall or 115ft fgjffhfjFjb

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Calculate the difference scores for the following data from a repeated measures study. Conduct a repeated measures t-test at apl

Delicious77 [7]

Answer:

There is no difference as per statistical evidence.

Step-by-step explanation:

We calculate t statistic from the formula

t =difference in means/Std error of difference

Here n1 = n2

t = (x bar - y bar)/sq rt of s1^2+s2^2

Let treatment I =X = 34 41 38 29

Treatment II Y = 39 48 35 36

X Y

Mean 35.50 39.50

Variance 81.00 105.00

H0: x bar = y bar

Ha: x bar not equal to y bar

(Two tailed test at 0.05 significant level)

N1 = 4 and N2 = 4

df=N1+N2-2 = 6

s1^2 = 81/3 =27 and s2^2 = 105/3 = 35

Std error for difference =

t = -1.02

p =0.348834

p>0.05

Since p value >alpha we accept null hypothesis.

Hence there is statistical evidence to show that there is no difference in the mean level of scores.

6 0

3 years ago

Can u help me on this... I will mark u as brilliant students.. plz

Colt1911 [192]

Answer:

For number 5 . a= 3 and b = -1

8 0

3 years ago

slavikrds [6]

Answer:

(24y^3x-20y^5x^2)/(-3y^5x^2)

= y^3x(24-20y^2x))/(-3y^5x^2)

=(24-20y^2x)/(-3y^2x)

=(24/(-3y^2x))-((20y^2x)/(-3y^2x))

=(-8/y^2x)-(-20/3)

=-8/y^2x + 20/3

7 0

2 years ago

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

$y=0.00035x^{2}$

Compute the first order derivative of <em>y</em> as follows:

$y=0.00035x^{2}$

$\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]$

$=2\cdot 0.00035x\\\\=0.0007x$

Now, it is provided that |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 605

Compute the arc length as follows:

$\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx$

$=\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\$

Now, let

$x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u$

$\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u$

$={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}$

Plug in the solved integrals in Arc Length and solve as follows:

$\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\$

$=1245.253707795227\\\\\approx 1245.25$

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

7 0

3 years ago

Answer?? what’s is the value of a

max2010maxim [7]

Answer:

13

Step-by-step explanation:

since the inside of a triangle is equal to 180

180-42-73=65

65/5=13 so 13 should be the answer.

6 0

3 years ago

Read 2 more answers