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4 years ago

Please help! Calculus: Derivative by limit process!!

Mathematics

1 answer:

ivolga24 [154]4 years ago

4 0

It doesn't necessarily have to be by the limit process but here you go:
A: 2x+1

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(01.02 LC) There are (36)2.30 candies in a store. What is the total number of candies in the store? (5 points)

EleoNora [17]

Answer:

82.8

Step-by-step explanation:

i think this is correct dont get mad if im wrong but im 80% sure im right

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3 years ago

kallie has a pottery business to make a certain piece she spends 3.68 on materials and 6.52 on labor. she then sells the pottery

Mrac [35]

Answer:

3/5 + 7/10 = 6/10 + 7/10 = 13/10 = 1 and 3/10 lbs. needed

She has 4/5 *3/1 bags = 12/5 = 2 and 2/5 pounds

2 2/5 - 1 3/10 = 2 4/10 - 1 3/10 = 1 and 1/10 pounds left over

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4 years ago

Jacqui has 12 gift baskets that each hold 15 small boxes of chocolate. Inside each small box, there are 17 pieces of chocolate,

denpristay [2]

Soo. If there are 12 gift baskets and 15 small boxes of chocolate in each box,

15*12 = 180 boxes of chocolate in total.

Since each box has 17 pieces of chocolate, there are 180 * 17 =3060 pieces of chocolate.

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3 years ago

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zepelin [54]

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3 years ago

It currently takes users a mean of 25 minutes to install the most popular computer program made by RodeTech, a software design c

ArbitrLikvidat [17]

Testing the hypothesis in this problem which is a two-tailed test, we can conclude that there is not sufficient evidence to conclude that the mean time of installation has changed, since the p-value of the test is 0.0364 > 0.01,

At the null hypothesis, we test if the mean is of 25 minutes, that is:

$H_0: \mu = 25$

At the alternative hypothesis, we test if the mean has changed, that is, if it is different than 25 minutes.

$H_1: \mu \neq 25$

We have the standard deviation for the sample, thus, the t-distribution is used. The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which:

X is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

In this problem:

25 is tested at the null hypothesis, thus $\mu = 25$ .
Sample mean of 26.2 minutes, thus $X = 26.2$ .
Sample of 51, thus $n = 51$ .
Variance of 16, thus $s = \sqrt{16} = 4$ .

The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}$

$t = 2.14$

The p-value of the test is found using a two-tailed test(test if the mean is different of a value), with t = 2.14 and 50 degrees of freedom.

Using a t-distribution calculator, this p-value is of 0.0364.

Since the p-value of the test is 0.0364 > 0.01, there is not sufficient evidence to conclude that the mean time of installation has changed.

A similar problem is given at brainly.com/question/23777908

3 0

3 years ago