I really need help on this please!!

the graph depicts Jacque's monthly supply of tuna. Suppose the price of tuna rises from $8 per round to $14 per round use the ar

Anna11 [10]

Answer:

Idk

Step-by-step explanation:

I’m dumb

7 0

3 years ago

Let x = 1 and y= -2, which expression has a negative value?

Nonamiya [84]

Answer:

The following expressions become negative:

1) x+y = 1+(-2) = -1

3) x times y = xy = (1)(-2) = -2

4) x divided by y = x/y = 1/-2 = -2

Step-by-step explanation:

Given

x = 1

y = -2

Computing by substituting x = 1, and y = -2

1) x+y = 1+(-2) = -1

2) x - y = 1 - (-2) = 1+2 = 3

3) x times y = xy = (1)(-2) = -2

4) x divided by y = x/y = 1/-2 = -2

It is clear from the above calculations that when we put x = 1, and y=-2 then the expressions (x+y), (x times y), and (x divided by y) become negative.

Therefore, the following expressions become negative:

1) x+y = 1+(-2) = -1

3) x times y = xy = (1)(-2) = -2

4) x divided by y = x/y = 1/-2 = -2

4 0

2 years ago

Complete the explanation of the error.

uysha [10]

Answer:

-2

Step-by-step explanation:

cause if you multiply a negative twice it becomes a positive

8 0

2 years ago

What is the relationship between area in the scale drawing and area on the on the actual tennis court if the scale is 1in.: 5ft.

sveticcg [70]

Given:

Length of the scale = 15.6 in.

Width of the scale = 7.2 in.

Scale of drawing = 1 in. : 5ft.

To find:

The ratio of area of the actual court to the area of the drawing (as a unit rate).

And to check whether it is the same as the ratio of length of the actual court to the length of the drawing.

Step-by-step explanation:

We have,

1 in. = 5ft.

Now, using this scale we get

15.6 in. = (15.6 × 5) ft =78 ft.

7.2 in. = (7.2 × 5) ft = 36 ft.

So, the actual length and width of tennis court are 78 ft and 36 ft respectively.

Area of actual tennis court is

$Area=length\times width$

$Area=78\times 36$

$Area=2808\text{ ft}^2$

The area of drawing is

$Area=15.6\times 7.2$

$Area=112.32\text{ in.}^2$

Now, ratio of area of the actual court to the area of the drawing (as a unit rate) is

$\dfrac{2808}{112.32}=\dfrac{25}{1}=25:1$

Ratio of area is 25:1 and ratio of length is 5:1 both area not same.

Therefore, the ratio of area of the actual court to the area of the drawing (as a unit rate) is 25:1.

7 0

3 years ago

Teacher raises A school system employs teachers at

Cerrena [4.2K]

By adding a constant value to every salary amount, the measures of

central tendency are increased by the amount, while the measures of

dispersion, remains the same

The correct responses are;

(a) The shape of the data remains the same

(b) The mean and median are increased by $1,000

(c) The standard deviation and interquartile range remain the same

Reasons:

The given parameters are;

Present teachers salary = Between $38,000 and $70,000

Amount of raise given to every teacher = $1,000

Required:

Effect of the raise on the following characteristics of the data

(a) Effect on the shape of distribution

The outline shape of the distribution will the same but higher by $1,000

(b) The mean of the data is given as follows;

$\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}$

Therefore, following an increase of $1,000, we have;

$\overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}$

$\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000$

Therefore, the new mean, is equal to the initial mean increased by 1,000

Median;

Given that all salaries, $x_i$ , are increased by $1,000, the median salary, $x_{med}$ , is also increased by $1,000

Therefore;

The correct response is that the median is increased by $1,000

(c) The standard deviation, σ, is given by $\sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}}$ ;

Where;

n = The number of teaches;

Given that, we have both a salary, $x_i$ , and the mean, $\overline x$ , increased by $1,000, we can write;

$\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}} = \sigma$

Therefore;

$\sigma_{new} = \sigma$ ; The standard deviation stays the same

Interquartile range;

The interquartile range, IQR = Q₃ - Q₁

New interquartile range, IQR $_{new}$ = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR

Therefore;

The interquartile range stays the same

Learn more here:

brainly.com/question/9995782

6 0

2 years ago