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2 years ago

How to write 25 thousands 56 ones

Mathematics

2 answers:

lilavasa [31]2 years ago

6 0

You would write it as 25,000 or it would be 0.025

Nadya [2.5K]2 years ago

3 0

Twenty five thousand fifty six

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(Right angle) Trigonometry please help!

kozerog [31]

Answer:

A = 41.4°

Step-by-step explanation:

Reference angle = A

Length of Adjacent side = 6

Length of Hypotenuse = 8

Apply the trigonometric function, CAH.

Cos A = Adj/Hyp

Substitute

$Cos(A) = \frac{6}{8}$

$A = Cos^{-1}(\frac{6}{8})$

A = 41.4° (approximated to the nearest tenth)

7 0

3 years ago

David is helping support a charity bike race. he is in charge of putting in a water station for the bikers halfway between the s

ryzh [129]

Answer:

A B C Or D

Step-by-step explanation:

7 0

2 years ago

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in in

Grace [21]

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

P.Q. = $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }$ = 5.063

$\sigma$ = population standard deviation

n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, $\sigma$ is ;

P(11.59 < $\chi^{2}__2_1$ < 32.67) = 0.90 {As the critical value of chi at 21 degrees

of freedom are 11.59 & 32.67}

P(11.59 < $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ < 32.67) = 0.90

P( $\frac{ 11.59}{(n-1) \times s^{2}}$ < $\frac{1}{\sigma^{2} }$ < $\frac{ 32.67}{(n-1) \times s^{2}}$ ) = 0.90

P( $\frac{(n-1) \times s^{2} }{32.67 }$ < $\sigma^{2}$ < $\frac{(n-1) \times s^{2} }{11.59 }$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1) \times s^{2} }{32.67 }$ , $\frac{(n-1) \times s^{2} }{11.59 }$ ]

= [ $\frac{21 \times 5.063^{2} }{32.67 }$ , $\frac{21 \times 5.063^{2} }{11.59 }$ ]

= [16.48 , 46.45]

90% confidence interval for $\sigma$ = [ $\sqrt{16.48}$ , $\sqrt{46.45}$ ]

= [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

5 0

3 years ago

30 POINTS: The art club had an election to select a president. 25% of the 76 members of the club voted in the election. How many

dimulka [17.4K]

Answer:

19 members voted.

Step-by-step explanation:

Percentage problems can be solved by a rule of three.

25% of the 76 members of the club voted in the election. How many members voted?

So 76 is 100% = 1. How much is 0.25?

76 - 1

x - 0.25

$x = 76*0.25$

$x = 19$

19 members voted.

7 0

3 years ago

In a certain normal distribution of scores, the mean is 50 and the standard deviation is 6. Find the z-score corresponding to a

Dovator [93]

Answer:

z-score = 1.667

Step-by-step explanation:

The provided information are:

Population mean $(\mu)$ = 50

Population standard deviation $(\sigma)$ = 6

Score (X) = 60

The formula to calculate the z-score is:

$z=\frac{X-\mu}{\sigma}$

The z- score can be calculated as:

$z=\frac{X-\mu}{\sigma}\\z=\frac{60-50}{6} \\z= 1.667$

Hence, the z-score is 1.667.

7 0

3 years ago