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3 years ago

In order to find the value of sin K in JKL, which of the following ratios needs to be used

Mathematics

1 answer:

nalin [4]3 years ago

3 0

Sin K= Opposide/Hypothenuse so you just go on from there using the proper triangle

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The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

3 years ago

Read 2 more answers

A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome

Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

$=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135$