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2 years ago

What is the unit rate $1.79 for 15 oz rounded to nearest tenth

Mathematics

1 answer:

ser-zykov [4K]2 years ago

8 0

Answer:

$0.119... per oz, or 0.1

Step-by-step explanation:

$.179 ÷ 15 = 0.119...

Now round to the nearest tenth, and you get 0.1.

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The diameters of cherry tomatoes produced by a large farm have an approximately Normal distribution, with a

Citrus2011 [14]

Answer: D. 0.9967

Step-by-step explanation: To solve this, you need to do the z-score formula with both numbers. So 30-22/2.5 = 3.2. On the z-score chart, that equals .9993. Hold onto that number. Then you do the same with 15. 15-22/2.5 = -2.8. On the z score chart, that equals .0026. Subtract those numbers. .9993 - .0026 = .9967. It's not as compacted as it looks <3

7 0

3 years ago

Hey I’m struggling with number 13 so please show your work. Thx

Dafna11 [192]

Answer:

$\frac{5}{12}$

Step-by-step explanation:

Because $\frac{3}{5}$ is being multiplied by S, we can divide $\frac{3}{5}$ from both sides of the equation. This will give us:

$\frac{1}{4}$ ÷ $\frac{3}{5}$

But, that looks a bit hectic. Instead of dividing, you can multiply by the reciprocal (which is essentially how to divide fractions). So, instead of $\frac{1}{4}$ ÷ $\frac{3}{5}$ , you get:

$\frac{1}{4}$ × $\frac{5}{3}$

When multiplying fractions, remember you can just multiply straight across-- numerator x numerator, then denominator x denominator.

By doing that, you get the fraction $\frac{5}{12}$

$\frac{5}{12}$ cannot be simplified any more, so S= $\frac{5}{12}$ is your answer :)

I hope this helps

6 0

3 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$