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3 years ago

Please help me with this :)

Mathematics

1 answer:

Elena L [17]3 years ago

8 0

Answer:

i think :

3 hours and 35 min

2 hours and 45 min

4 hours and 30 min

4 hours and 5 min

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How to evaluate 4(x+2−5x) when x=2

Crank

Answer:

-24

Step-by-step explanation:

4(x+2−5x)

Combine like terms

4(2 -4x)

Distribute

8 -16x

Let x=2

8 - 16(2)

8 - 32

-24

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3 years ago

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8.5 written as a decimal??

Galina-37 [17]

Its already written as a decimal but it can be 8 1/2

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4 years ago

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Of 450 college students, 110 are enrolled in math, 205 are enrolled in English, and 50 are enrolled in both. If a student is sel

likoan [24]

Answer:

Step-by-step explanation:

Given that Of 450 college students, 110 are enrolled in math, 205 are enrolled in English, and 50 are enrolled in both. If a student is selected at random

the probability

(a) The student is enrolled in mathematics= $\frac{110}{450} =0.244\\$

(b) The student is enrolled in English.= $\frac{205}{450} =0.456\\$

(d) The student is enrolled in mathematics or English.= $\frac{110+205-50}{450} =0.589\\$

(e) The student is enrolled in English but not in mathematics.

= $\frac{205-50}{450} =0.344\\$

(f) The student is not enrolled in English or is enrolled in mathematics.

= $\frac{450-205+110}{450} =0.7889\\$

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3 years ago

Please help me with questions 1 and 2

Nana76 [90]

1. 3/3 2. 3/4 the answer is 3/4

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3 years ago

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NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud

SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

$M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }$

The 85% confidence level $z_{\alpha } = 1.44$

$\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}$

$\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}$

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

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3 years ago