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2 years ago

Find the valie of V when M=2 and E =64 when V= + 2E/m

Mathematics

2 answers:

BabaBlast [244]2 years ago

8 0

Answer:

+ 64

Step-by-step explanation:

Given,

M = 2

E = 64

To find : V = ?

V = + 2E/M

V = + 2 x 64 / 2

= + 2 x 32

V = + 64

Ivenika [448]2 years ago

7 0

Answer:
132.
Division:
264 / 2 = 132.
Fraction:
264/2 = 132.
Hope this helps.

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Solve for C. 4/ 7(B - C) = A C = ___

klio [65]

Answer:

C = ¼(4B – 7A)

Step-by-step explanation:

⁴/₇(B – C) = A Multiply each side by 7

4(B - C) = 7A Remove parentheses

4B – 4C = 7A Subtract 4B from each side

-4C = 7A – 4B Multiply each side by -1

4C = -7A + 4B Divide each side by 4

C = ¼(4B – 7A)

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3 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

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Using words, explain all steps of dimensional analysis when converting 6.7 hectometers to meters.

9966 [12]

<h2>6.7 hectometers is equal to "670 meters".</h2>

Step-by-step explanation:

We have,

6.7 hectometers

To convert 6.7 hectometers to meters.

We know that,

1 hectometer = 100 meters

⇒ 1 hm = 100 m

∴ 6.7 hectometers

= 6.7 × 100 meters

= 670 meters

∴ 6.7 hectometers (hm) = 670 meters (m)

Thus, 6.7 hectometers is equal to "670 meters".

3 0

3 years ago

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar

Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ounces, with a sample standard deviation of 0.052 ounces.

Let $\mu$ = population mean amount packaged.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.17 ounces {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 8.17 ounces {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount dispensed = 8.171 ounces

s = sample standard deviation = 0.052 ounces

n = sample of packages = 50

So, the test statistics = $\frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }$ ~ $t_4_9$

= 0.1359

The value of t-test statistics is 0.1359.

Also, the P-value of test-statistics is given by;

the meaning of the p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

P-value = P( $t_4_9$ > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

6 0

3 years ago