If x=a sintheta and y=bcostheta, show that (b^2)(x^2)+(a^2)(y^2)=(a^2)(b^2)a

A waffle house offers 19 different types of syrups. If a customer can choose any number of syrups to have with their waffle, how

bogdanovich [222]

Answer:

Step-by-step explanation:

they all just have 19 or less

3 0

3 years ago

In March, you have $220 in your savings account. By the end of April, your savings have increased by 30%. How much money do you

Lana71 [14]

Answer:

$286

Step-by-step explanation:

220:100=2.2

2.2 x 30 = 66

220+66=286

6 0

3 years ago

What is the LCM of 37 and 12?

artcher [175]

$37|37 \\ 1 \\ \\ 12|2 \\ 6 \ |2 \\ 3 \ |3 \\ 1 \\ \\ 12=2^2*3 \\ \\ L.C.M(37;12)=444$

6 0

4 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

A school has 2 computer labs. Each lab has 30 computers. A total of

Korolek [52]

Answer:

(2*30)-6

Step-by-step explanation:

8 0

3 years ago

If x=a sintheta and y=bcostheta, show that (b^2)(x^2)+(a^2)(y^2)=(a^2)(b^2)a​

If x=a sintheta and y=bcostheta, show that (b^2)(x^2)+(a^2)(y^2)=(a^2)(b^2)a