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3 years ago

There are 32 cookies in a 20-ounce package of Oreos. How many Oreos are in a 15-ounce package?

Mathematics

2 answers:

Helga [31]3 years ago

4 0

Answer:

2.133

Step-by-step explanation:

Hope this helps!

Ipatiy [6.2K]3 years ago

3 0

Answer:

It's 24 :D

Step-by-step explanation:

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The Price of a machine depreciates at the rate of ₹1500 per year. If it was bought for ₹15,800, what will be its price after 5 y

saul85 [17]

Answer:

5th year= 9800

Step-by-step explanation:

1st year=15800

2nd=14300

3rd=12800

4th=11300

5th=9800

8 0

3 years ago

how many pounds of rock will you need to fill an area 25 ft by 35 ft to a depth of 2 in. If the rock weighs 1050 lb cubic yard?

Eduardwww [97]

The volume of the area is Length x width x height:

25 x 35 x 2/12 = 145.83 cubic feet.

1 cubic foot = 0.037 cubic yards:

145.83 x 0.037 = 5.40 cubic yards.

5.4 cubic yards x 1050 lbs/ cubic yard = 5,665.6 pounds.

Note:

Rounding and conversion may change the answer slightly, but I believe I rounded everything as it should be.

6 0

3 years ago

What is -4(7x+5)=-160

puteri [66]

Answer is -23

Add 4 from both sides

Subtract 5 from both sides

Then divide 7

6 0

3 years ago

Read 2 more answers

Find all y such that the distance between the points (8,10) and (4,y) is 20.

andriy [413]

Answer:

y = 394

Step-by-step explanation:

8 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