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3 years ago

LOOK AT CHART BEFORE THE BULLET PART:

Mathematics

1 answer:

Ber [7]3 years ago

5 0

Answer:

1st blank: $\text{Slope(m)}=-\frac{3}{4}$

2nd blank: $\text{y-intercept}=399$

3rd blank: Slope-intercept form: $y=-\frac{3}{4}x+399$

Step-by-step explanation:

Let x be the number of sandwiches and y be the number of wraps.

We have been given a chart of values, which represents Sal's total profit on lunch specials for two months. We are asked to fill in the empty boxes.

We have been given that Sal gets a profit of $3 after selling each sandwich, so Sal's profit after selling x sandwiches will be 3x.

As each wrap gives a profit of $4, So Sal's profit after selling y wraps will 4y.

Since Sal’s total profit on lunch specials for next month is $1596. We can represent this information in an equation as:

$3x+4y=1596$

We can see that our equation is in standard form, so let us convert it in slope-intercept form of equation.

Since we know that equation of a line in slope-intercept form is: $y=mx+b$ , where,

m = Slope of line,

b = y-intercept or initial value.

Let us subtract 3x from both sides of our equation.

$3x-3x+4y=1596-3x$

$4y=1596-3x$

Let us divide both sides of our equation by 4.

$\frac{4y}{4}=\frac{1596-3x}{4}$

$y=\frac{1596}{4}-\fraxc{3x}{4}$

$y=399-\frac{3}{4}x$

$y=-\frac{3}{4}x+399$

Therefore, equation $y=-\frac{3}{4}x+399$ represents the Sal's total profit for 2nd month in slope-intercept form of equation.

Upon comparing our equation with slope-intercept form of equation we can see that slope of our line is $-\frac{3}{4}$ and y-intercept is 399.

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Standing on a cliff 380 meters above the sea, Pat sees an approaching ship and measures its angle of depression, obtaining 9 deg

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Answer:

(a). 2399.23 meters.

(b). 1944.19 meters apart.

Step-by-step explanation:

Please find the attachment.

We have been given that standing on a cliff 380 meters above the sea, Pat sees an approaching ship and measures its angle of depression, obtaining 9 degrees.

(a) We are asked to find how far from the shore is the ship.

We know that tangent relates opposite side of a right triangle with its adjacent side.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(9^{\circ})=\frac{380}{x}$

$x=\frac{380}{\text{tan}(9^{\circ})}$

$x=\frac{380}{0.158384440325}=2399.22557$

Therefore, the ship is approximately 2399.23 meters away from the shore.

$\text{tan}(5^{\circ})=\frac{380}{x+y}$

$x+y=\frac{380}{\text{tan}(5^{\circ})}$

$x+y=\frac{380}{0.087488663526}$

$2399.22557+y=4343.419875$

$y=4343.419875-2399.22557$

$y=1944.194305\approx 1944.19$

Therefore, the both ships are approximately 1944.19 meters apart.

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3 years ago

Arthur made the Scatterplot below of the temperature outside his house over the course of nine days at the begging of winter if

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4 years ago

The function f is continuous on the interval [3, 13] with selected values of x and f(x) given in the table below. Find the avera

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The average rate of change of $f(x)$ over the interval [3, 13] is given by the ratio

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3 years ago

A coin is tossed 5 times. Find the probability that exactly 1 is a tail. Find the probability that at most 2 are tails.

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Answer:

Step-by-step explanation:

<h2>First question</h2>

The only possibilities where there is exactly 1 tail are:

(t,h,h,h,h)
(h,t,h,h,h)
(h,h,t,h,h)
(h,h,h,t,h)
(h,h,h,h,t)

those are 5 favorable outcomes.

where <em>h represent heads</em> and <em>t represent tails. </em>There are $2^5 32$ total number of outcomes after tossing the coin 5 times. Because every time you toss the coin, you have 2 possibilities, and as you do it 5 times, those are $2^5$ options. We can conclude from this that

The probability that exactly 1 is a tail is $5/32$ .

<h2>Second question</h2>

We already know the total number of outcomes; 32. Now we need to find the number of favorable outcomes. In order to do that, we can divide our search in three cases: <em>1.-there are no tails, 2.-exactly 1 is a tail, 3.- exactly 2 are tails.</em>

The first case is 1 when every coin is a head. The second case we already solved it, and there are 5. The third case is the interesting one, we can count the outcomes as we did in the previous questions, but that's only because there are not too many outcomes. Instead we are going to use combinations:

We need to have <u>2</u> tails, the other coins are going to be heads. We made <u>5</u> tosses, then the possible combinations are $C_{5,2} = \frac{5!}{3!2!} = \frac{120}{6*2} = 10$

Finally, we conclude that there are 1 + 5 + 10 favorable outcomes, and this implies that

The probability that at most 2 are tails is $\frac{16}{32} = \frac{1}{2}$ .

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