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3 years ago

HELP WANTED with this equation!

Mathematics

1 answer:

dmitriy555 [2]3 years ago

3 0

Answer:

Step-by-step explanation:

5/8 = 15/24 times 4 2.5 an hour

2/3 = 16/24 times 5 3 an hour

Machine B has the greater rate

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A triangle has a base length of 12 inches and a height of 14 inches. From that

Paladinen [302]

A= bh(1/2)
a=12(14)(1/2)
a= 168(1/2)
a= 84inches

3 0

3 years ago

Please help me with my question,Thanks!! (By the way the ratio is 12 to 36)

saul85 [17]

To answer this, you would use the original ratio of 2 to 5 and create an equivalent ratio that includes the 12 tulips. You would get a ratio of 12 to 30. This is not enough daisies.

I then tried 2/5 times 7/7, but it still did not reach the 36 daisies that are there. When you use a factor of 8, you will get 16 tulips to 40 daisies. This works!

You would need to add 4 tulips and 4 daisies to what he already has.

5 0

4 years ago

7. A number of minor automobile accidents occur at various high-risk intersections in Teton County despite traffic lights. The T

Anit [1.1K]

Answer:

Modification has not reduced the number of accidents.

Step-by-step explanation:

$H_0: \mu\geq0\\H_a:\mu$

A B C D E F G H

Before modification 5 7 6 4 8 9 8 10

After modification 3 7 7 0 4 6 8 2

Difference -2 0 1 -4 -4 -3 0 -8

Mean of differences $M_d = \frac{sum}{8}=\frac{2+0-1+4+4+3-0+8}{8}=2.5$

Standard deviation of differences = $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}=2.9277$

Formula of paired t test :

$t=\frac{M_d}{\frac{s}{\sqrt{n}}}\\t=\frac{2.5}{\frac{2.9277}{\sqrt{8}}}\\t = 2.4152$

Df = n-1 = 8-1 =7

$t critical = t_{(df, \alpha)}=t_{7,0.01}=2.998$

t critical> t calculated

So, We failed to reject null hypothesis

Hence modification has not reduced the number of accidents.

6 0

3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

3 years ago

Which equation can be used to represent the statement?

Luba_88 [7]

Answer:

4. One-sixth n minus 2 = negative two-thirds

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers