Ask question

Ask question

All categories

aleksandr82 [10.1K]

3 years ago

15

during the week. Jimmy spent $45 eating out, gain $55 mowing lawns and spent $15 on gas. What was Jimmy’s change in dollars at t

he end of the week?

2 answers:

Veronika [31]3 years ago

8 0

Answer:

5

Step-by-step explanation:

Let's look at the question.

Spent- He spent 45 dollars eating out.

He spent an additional 15 dollars on gas

Earned- He earned 55 dollars from mowing lawns

Earned is going to be a positive value.

Spent is going to be a negative value.

The situations don't matter in the context of the question. They are just asking how much he got charged.

The amount charged is always positive. You can never have a negative charged amount of money. That would mean the bank owed you money, which would never happen.

Let's substitute the values into an equation.

The <em>variable C</em> will stand for amount charged.

C = -45 + 55 - 15

Combine the negative values to get -60

Add -60 to 55.

You will get -5, which then you will have to take the absolute value of.

The absolute value will be the charge in dollars at the end of the week.

jeka57 [31]3 years ago

6 0

Answer:

-$5

Step-by-step explanation:

-$45+$55=$10

$10-$15=-$5

You might be interested in

Solve the equation sin x + cos x=cos 2x for 0 27. x

vodka [1.7K]

Answer:

$x=\frac{\pi}{4}$ and $x=-\frac{3\pi}{4}$

Step-by-step explanation:

We are given that sin x+cos x=cos 2x

We have to solve the given equation for $0\leq x\leq 2\pi$

$sin x+cos x=cos^2x-sin^2x$

Because $cos 2x=cos^2-sin^2$

$sinx+cos x=(sinx +cos x)(sinx-cos x)$

$1 =sin x-cos x$

$sin x=cos x$

$\frac{sinx }{cos x}=1$

$tan x=1$

$tan x=\frac{sinx}{cos x}$

$tan x=tan\frac{\pi}{4}$

$x=\frac{\pi}{4}$

Tan x is positive in I and III quadrant

In III quadrant angle $\theta$ replace by $\theta -\pi$

Therefore, $tan x=tan (\frac{\pi}{4}-\pi)=tan\frac{\pi-4\pi}{4}=tan\frac{-3\pi}{4}$

$x=-\frac{3\pi}{4}$

8 0

3 years ago

6x to the power of 2 + 5x +3(2x to the power of 2) + 7x

IRINA_888 [86]

Answer: 6x² + 5x + 3 • 2x² + 7x = 12x² + 12x

Explanation:

6x² + 5x + 3 • 2x² + 7x

Multiply 3 by 2 to get 6. (6x² + 5x + 6x² + 7x)

Add 6x² and 6x² to get 12x². (12x² + 5x + 7x)

Add 5x and 7x to get 12x. (12x² + 12x)

6 0

2 years ago

Can somebody please give me a detailed answer

Serjik [45]

Lets go ahead and calculate the area of each side; added together, this will give us the full surface area. First the base, which is 9 by 9 and a square. This means its area is 9*9, which equals 81. Next the triangles, which are all the same, so we only have to do it once and then add it 4 times at the end. The area of a triangle is (1/2)bh, so (1/2)(9)(7) = 31.5.

Finally, we just need to add the areas up.
81 + 31.5 + 31.5 + 31.5 + 31.5 = 81 + 4(31.5)
81 + 126 = 207 units^2

3 0

3 years ago

Solve 5x 2 - 7x + 2 = 0 by completing the square. What are the solutions?

Alexeev081 [22]

Answer:

$x=\frac{2}{5}$ or $x=1$

Step-by-step explanation:

The given quadratic equation is

$5x^2-7x+2=0$

Group the constant terms on the right hand side.

$5x^2-7x=0-2$

$5x^2-7x=-2$

Divide through by 5.

$x^2-\frac{7}{5}x=-\frac{2}{5}$

Add the square of half the coefficient of x., which is $(\frac{1}{2}\times- \frac{7}{5})^2=\frac{49}{100}$ to both sides of the equation.

$x^2-\frac{7}{5}x+\frac{49}{100}=-\frac{2}{5}+\frac{49}{100}$

The left hand side is now a perfect square.

$(x-\frac{7}{10})^2=\frac{9}{100}$

Take the square root of both sides;

$(x-\frac{7}{10})=\pm \sqrt{\frac{9}{100}}$

$x-\frac{7}{10}=\pm \frac{3}{10}$

$x=\frac{7}{10}\pm \frac{3}{10}$

$x=\frac{7-3}{10}$ or $x=\frac{7+3}{10}$

$x=\frac{4}{10}$ or $x=\frac{10}{10}$

$x=1$ or $x=\frac{2}{5}$

5 0

3 years ago

Read 2 more answers

Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba

stiv31 [10]

Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= $\frac{1}{n-1}$ *[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= $\frac{1}{11}$ *[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= $\frac{1}{9}$ *[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

$F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}$

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

$F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10$

The rejection region is then F ≥ 3.10

$F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006$

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

8 0

3 years ago