What is the answer to 200 x 32.6

A plane is flying at 32, 460 feet. To avoid some turbulence, the plane ascends 1645 feet. It then descends 3/5 of its new height

aleksley [76]

Answer: C and E

Step-by-step explanation:

The plane starts at 32,460 feet.

It then ascends 1645 feet. Ascending means it's going up higher. So we add.

32,460 + 1645 = 34,105 (C)

From here (34,105 feet) the plane descends by 3/5 of it's height. Descending means it's lowering. So we subtract.

But first we have to figure out how much we are subtracting.

3/5 is equal to .6

Multiply to find 3/5 of current height.

34,105 × .6 = 20,463

So the plane descends by 20,463 feet.

34,105 - 20,463 = 13,642 feet (E)

3 0

4 years ago

Joe has read 30% of a book. He has 49 more pages to finish. How many pages are there in the book?

sasho [114]

Answer:

70 pages

Step-by-step explanation:

100%-30%=70%

70% of the book is 49 pages.

x is the total number of pages.

0.7x=49

7x=490

x=70

7 0

3 years ago

Solve the following expressions using logarithm and Antilogarithm .

katrin [286]

Answer:

0.01688496
0.3449537
0.05002308
2.0375623
0.4162862

Step-by-step explanation:

Some formulas to help

log ab = log a + log b
log a/b = log a - log b
log a^b = b log a
antilog (log a) = a, antilog is the inverse of log
get values of log and antilog by using calculator or online calculator ( I used online calculator for this problem)
round numbers as required, I left them as is

1. Let the number be x, solving to show the method

√0.0002851 = x
log √0.0002851 = log x
1/2 log 0.0002851 = log x
1/2(-3.545) = log x
log x = -1.7725
antilog (log x) = antilog (-1.7725)
x = 0.01688496

2. Short of the above method, will apply to this and following

$\sqrt[7]{0.0005812}$ =
antilog (1/7 log (0.0005812)) =
antilog (1/7(-3.23567439437)) =
antilog (-0.46223919919) =
0.3449537

3. .........................................

2.714^3 =
antilog (log 2.714^3) =
antilog (3 log 2.714) =
antilog (3*0.43360984332) =
antilog (1.30082952996) =
0.05002308

4..........................................

35.12^(1/5) =
antilog (1/5 log (35.12)) =
antilog (1/5*1.54555450723)
antilog (0.30911090144) =
2.0375623

5. .........................................

(0.07214)^(1/3) =
antilog ( 1/3 log (0.07214)) =
antilog (1/3*(-1.14182386202 )) =
antilog ( --0.380607954 )=
0.4162862

Let me know if anything is not clear. Hope it helps.

5 0

3 years ago

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bagirrra123 [75]

hope it's correct answer.....and it helps

3 0

3 years ago

In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is

VladimirAG [237]

Answer:

Null hypothesis is $\mathbf {H_o: \mu > 937}$

Alternative hypothesis is $\mathbf {H_a: \mu < 937}$

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Step-by-step explanation:

Given that:

Mean $\overline x$ = 960 hours

Sample size n = 36

Mean population $\mu =$ 937

Standard deviation $\sigma$ = 52

Given that the mean time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is $\mathbf {H_o: \mu > 937}$

Alternative hypothesis is $\mathbf {H_a: \mu < 937}$

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

$Z = \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}$

$Z = \dfrac{960-937 }{\dfrac{52}{\sqrt{36}}}$

$Z = \dfrac{23}{8.66}$

Z = 2.65

The decision rule is to reject null hypothesis if test statistics is greater than critical value.

CONCLUSION:

Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975 (using the Excel Function: =NORMDIST(z))

= 0.004025

6 0

3 years ago