What is the answer to this ???

Answer all five questions below

Svetach [21]

Answer:

use a calculator

Step-by-step explanation:

there for question b

6 0

3 years ago

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime

Citrus2011 [14]

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2$

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

$M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46$

The standard error is

$s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194$

The t-statistic is

$t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52$

The degrees of freedom are

$df=n_1+n_2-2=9+9-2=16$

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

5 0

2 years ago

Work out the formula of the nth term in the following quadratic sequence:<br> 19, 15, 9, 1...

Butoxors [25]

In a quadratic sequence we'll get a linear first difference and a constant second difference. Let's verify that.

n 1 2 3 4

f(n) 19 15 9 1

1st diff -4 -6 -8

2nd diff 2 2

We see that we got a constant second difference. We could just extend that and work back up to get more values.

n 1 2 3 4 5 6 7

f(n) 19 15 9 1 -9 -21 -35

1st diff -4 -6 -8 -10 -12 -14

2nd diff 2 2 2 2 2

That's just an aside; we're after the general formula. We have

f(1)=19, f(2)=15, f(3)=9

In general we can assume

f(n) = an² + bn + c

We get three equations in three unknowns,

19 = a(1²)+b(1)+c = a+b+c

15 = a(2²) + b(2) + c = 4a + 2b + c

9 = a(3²) + b(3) + c = 9a + 3b + c

That's a 3x3 linear system; it's easy to solve directly. Subtracting pairs,

4 = -3a - b

6 = -5a - b

Subtracting those,

-2 = 2a

a = -1

b = -3a -4 = -1

c = 19-a-b = 21

Answer: f(n) = -n² - n + 21

Check:

f(1) = -1 - 1 + 21 = 19, good

f(2) = -4 - 2 + 21 = 15, good

f(3) = -9 - 3 + 21 = 9, good

f(4) = -16 - 4 + 21 = 1, good

Let's check our extended table, how about

f(7)= -49 - 7 + 21 = -35, good

6 0

3 years ago

There are 36 students in Mrs.Keatons's sixth grade class. If 5/12 of her students are girls, how many girls in the class?

murzikaleks [220]

There are 15 girls because 12 x 3 = 36 and 5 x 3 = 15

6 0

3 years ago

connie sells lemonade for $0.25 per small cup and $0.75 per large cup. if she sells 20 small cups how many large ones must she s

Nady [450]

Answer:34

Step-by-step explanation:

8 0

3 years ago