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3 years ago

6

Monty has a total of $290 in ten dollar and five dollar bills. This can be represented by the function 10x + 5y = 290. Interpret

the x- and y-intercepts.

1 answer:

statuscvo [17]3 years ago

5 0

Answer:

The x-intercept indicates that he has 29 ten dollar bills

and no five dollar bills. The y-intercept indicates that he has 58 five

dollar bills and no ten dollar bills.

Step-by-step explanation:

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A store sells a box of high liters that contains 4 yellow, 3 blue, 2 pink, and 1 green high liter. What is the probability of ra

Iteru [2.4K]

There are 10 in total
3/10 x 2/9 = 6/90
Reduce
1/15
1/15 is the chance of that happening

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3 years ago

Help please! I have to do this math test and I dont understand

n200080 [17]

Answer:

y=1x+1

Step-by-step explanation:

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4 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

3 years ago

Please help me! I need this answered fast!

Sonja [21]

Answer:

1. I think it's A.

2. I think it's D

8 0

3 years ago

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t)

beks73 [17]

H(t) = Ho +Vot - gt^2/2

Vo = 19.6 m/s
Ho = 58.8 m
g = 9.8 m/s^2

H(t) = 58.8 + 19.6t -9.8t^2/2 = 58.8 + 19.6t - 4.9t^2

Maximun height is at the vertex of the parabole

To find the vertex, first find the roots.

58.8 + 19.6t - 4.9t^2 = 0

Divide by 4.9

12 + 4t - t^2 = 0

Change sign and reorder

t^2 - 4t -12 = 0

Factor

(t - 6)(t + 2) =0 ==> t = 6 and t = -2.

The vertex is in the mid point between both roots

Find H(t) for: t = [6 - 2]/2 =4/2 = 2

Find H(t) for t = 2

H(6) = 58.8 + 19.6(2) - 4.9(2)^2 = 78.4

Answer: the maximum height is 78.4 m

5 0

3 years ago