All categories

3 years ago

a piece of wood weighing 10 ounces is found to have a weight of 8 ounces after drying. the moisture content was:

1 answer:

8 0

Moisture content should be the difference of the weight of the object wet (10 oz.) and the weight of the object when it was dry(8 oz.)
So,
10 oz. - 8 oz. =moisture content, or 2 oz.

You might be interested in

ANSWER PLEASE *50 POINTS*

zvonat [6]

If a football player runs 5/6 of a mile every 3/4th of a hour, that means every 1/4th a hour he runs 1.5/6th of a mile.

That means every 15 minutes the football player runs .2 of a mile. That means that every hour he runs :

0.8 of a mile!

3 0

3 years ago

If N = f(x) and f(x) = 3x^2 −x−5

harina [27]

Answer:

I really don't know so sorry

8 0

3 years ago

Problem Page Maya purchased a prepaid phone card for $30 . Long distance calls cost 18 cents a minute using this card. Maya used

Helen [10]

Her call lasted 17minutes

8 0

2 years ago

A square pizza box has an area of 121 in 2 What is the length of each of its sides?

Anuta_ua [19.1K]

Answer:

11 in

Step-by-step explanation:

Area of a square is L^2

where L is length.

L^2 = 121 in^2

FIND THE SQUARE ROOT OF BOTH SIDES (including in^2).

L= 11 in.

side = 11 in

Please give brainliest, like, and rate 5 stars. I hope this helps. Thank you.

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago