Add 5 5 10 15 and add 3 3 6 9 what are tje first four ordered pairs formed from corresponding terms of the two numbers

SAT verbal scores are normally distributed with a mean of 433 and a standard deviation of 90. Use the Empirical Rule to determin

laila [671]

34% of the scores lie between 433 and 523.

Solution:

Given data:

Mean (μ) = 433

Standard deviation (σ) = 90

<u>Empirical rule to determine the percent:</u>

(1) About 68% of all the values lie within 1 standard deviation of the mean.

(2) About 95% of all the values lie within 2 standard deviations of the mean.

(3) About 99.7% of all the values lie within 3 standard deviations of the mean.

$$Z(X)=\frac{x-\mu}{\sigma}$

$$Z(433)=\frac{433-\ 433}{90}=0$

$$Z(523)=\frac{523-\ 433}{90}=1$

Z lies between o and 1.

P(433 < x < 523) = P(0 < Z < 1)

μ = 433 and μ + σ = 433 + 90 = 523

Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.

i. e. $((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%$

Here μ to μ + σ = $\frac{68\%}{2} =34\%$

Hence 34% of the scores lie between 433 and 523.

8 0

3 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago

2sec^2x - secx - 1 = 0

Ivenika [448]

What are you trying to do here?
Solve the graph, or make it appear as something else?

First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0

sec (x) (2sec (x) -2) = 0

Then we're going to separate the two to find the zeros of each because anything time 0 is zero.

sec(x) = 0

2sec (x) - 2 = 0

Now, let's simplify the second one as the first one is already.

Add 2 to both sides:

2sec (x) = 2

Divide by 3 on both sides:

sec (x) = 1

I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!

8 0

3 years ago

The olympics are held every four years. In 2016, the XXXI olympiad were held in Rio de Janeiro,Brazil. Determine the year of the

Anna007 [38]

Answer:

First Olympic held in year 1896.

Step-by-step explanation:

Number of Olympiad held in Rio = XXXI = 31st

Difference of years in two consecutive Olympiads = 4 years

Number of years spent in 31 Olympiads can be calculated by,

Number of years spent = (n - 1)d

Here n = number of Olympiads held

d = difference between two consecutive Olympiads

Number of years spent till 31st Olympiads = (31 - 1)×4

= 120 years

Therefore, 1st Olympiad held in the year = 2016 - 120

= 1896

6 0

2 years ago

200 to 170 what is the percentage discount offered

wel

Answer:

85%

Step-by-step explanation:

Convert fraction (ratio) 170 / 200 Answer: 85%

6 0

3 years ago