All categories

3 years ago

1 ton is equal to 2000 pounds.how many pounds are in 4 tons, 568 pounds?

Mathematics

2 answers:

LekaFEV [45]3 years ago

6 0

Answer:

Hello There!!

Step-by-step explanation:

If 1 ton=2000 (1 ton to 4 is times by 4 so 2000

4 tons=8000 is times by 4)

hope this helps, have a great day!!

~Pinky~

xz_007 [3.2K]3 years ago

6 0

$\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}$

Ur answer is

8568 pounds.

Thanks
Hope it helps.

You might be interested in

Write the numbers -2.1, -1 5/6, -1 2/3 and -2.2 in the form a/b from greatest to least

Airida [17]

Answer:

-1 5/6, 1 2/3, -2.1, -2.2

Step-by-step explanation:

Rember the numbers are neggative, then convert -1 2/3 to -1 4/6.

4 0

3 years ago

You have decided to purchase a car for $22,346.16. The credit union requires a 10% down payment and will finance the

sukhopar [10]

Answer:

(b) $\$24,169.60$

Step-by-step explanation:

Given: Cost of car is $\$22,346.16$

Sales tax is $7.6\%$

The license and title charges are $\$125.13$

To find: The amount that the credit union will finance.

Solution:

We have,

Cost of car $=\$22,346.16$

Sales tax is $7.6\%$ of $\$22,346.16$

Therefore,

sales tax $=\frac{7.6}{100}\times22,346.16=\$1698.31$

The license and title charges $=\$125.13$

Now, the amount that the credit union will finance is

cost of car $+$ sales tax $+$ the license and title charges

$=\$22346.16+\$1698.31+\$125.13$

$=\$24,169.60$

Hence, the amount that the credit union will finance is $\$24,169.60$ .

So, (b) $\$24,169.60$ is correct.

5 0

3 years ago

Read 2 more answers

To find proportion of the area under the normal curve between two Z scores that are both above the mean, it is necessary to exam

VikaD [51]

It is necessary to imagine the sum of the areas between each z-score and the average.

Given as the ratio of the area under the normal curve between two z-scores, both above average.

The Z score accurately measures the number of standard deviations above or below the mean of the data points.

The formula for calculating the z-score is

z = (data points – mean) / (standard deviation).

It is also expressed as z = (x-μ) / σ.

A positive z-score indicates that the data points are above average.
A negative z-score indicates that the data points are below average.
A z-score close to 0 means that the data points are close to average.
The normal curve is symmetric with respect to the mean and needs to be investigated.

Therefore, to find the percentage of the area under the normal curve between two z-scores, both above the mean, you need to look at the sum of the areas between the z-score and the mean.

Learn more about z-score from here brainly.com/question/16768891

#SPJ4

7 0

1 year ago

What expression is equivalent to 6+3(n-4)-8+2n

lorasvet [3.4K]

Answer:

Step-by-step explanation:

Comment

Begin by removing the brackets.

6+3(n-4)-8+2n

6 + 3n - 12 - 8 + 2n Collect like terms

3n + 2n + 6 - 12 - 8 Combine

5n - 14

Answer

5n - 14

4 0

2 years ago

A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th

ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\$

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

$H_0: \mu=500\\\\H_a:\mu< 500$

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

4 0

3 years ago