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2 years ago

A submarine 50.2 m below the surface of the ocean

Mathematics

1 answer:

Harman [31]2 years ago

8 0

Answer:

101.5 m

Step-by-step explanation:

What you have to do is 50.2 + 15.6 which equals 65.8, and then you add another 35.7 and the answer would be 101.5 m

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Two weeks in a row, the golf course hosts a group of golfers. The second week had 10 more golfers than the first week. Use the d

Trava [24]

Answer:

Option A.

Step-by-step explanation:

From the first dot plot the given data set for week 1 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69), (70, 70, 72, 72, 72), (75, 75, 76, 78, 78), (80, 80, 80, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{72+75}{2}=73.5$

$Mean=\frac{\sum x}{n}=\frac{1491}{20}=74.55$

From the second dot plot the given data set for week 2 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 85, 86, 86, 86, 88, 88, 88, 88, 89, 89, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69, 70, 70), 72, (72, 72, 75, 75, 76, 78, 78), (80, 80, 80, 85, 86, 86, 86), 88, (88, 88, 88, 89, 89, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{78+80}{2}=79$

$Mean=\frac{\sum x}{n}=\frac{2364}{30}=78.8$

The range for Week 1 is equal to the range for Week 2.

The median for Week 2 is more than the median for Week 1.

The mean for Week 2 is more than the mean for Week 1.

Therefore, the correct option is A.

6 0

2 years ago

Daniel is creating a scale drawing of her room. the rectangular room measures20 1/2 ft. by 25 ft. if her drawing uses the scale

zavuch27 [327]

1 in = 2 ft

20.5 ft = 10.25 in
25 ft = 17.5 in

8.5 in and 11 in. no, the drawing will not fit, because the proportions is larger than 8.5 in and 11 in.

hope this helps

5 0

2 years ago

Read 2 more answers

Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ

MrRissso [65]

Answer:

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

Step-by-step explanation:

We get the limits of integration:

$R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq 4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace$

We use the spherical coordinates and we calculate a triple integral:

$V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4 \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\$

we get:

$V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}$

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

4 0

2 years ago

Write the equation of a circle with center at (3, -2), which also passes through the point (0, 2).

Airida [17]

$\text{The equation of acircle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\(h;\ k)-center\\r-radius\\\\\text{We have center}\ (3,\ -2)\to h=3\ \text{and}\ k=-2.\\\\\text{The radius is the distance between the points} (3,\ -2)\ \text{and}\ (0,\ 2).\\\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{Substitute:}\\\\d=\sqrt{(2-(-2))^2+(0-3)^2}=\sqrt{4^2+(-3)^2}=\sqrt{16+9}\\\\=\sqrt{25}=5\\\\Answer:\ \boxed{(x-3)^2+(y-(-2))^2=5^2\to\boxed{(x-3)^2+(y+2)^2=25}}$

3 0

3 years ago

Micah has 294 songs stored in his phone,which is 70% of the songs that Jorge in his phone.How many songs are stored on Jorge'S p

Ipatiy [6.2K]

Answer:

206

Step-by-step explanation:

70% of 294

70/100 × 294/1

205.8

round it off to 206

8 0

2 years ago