Help me again pleasee

iii). Sheridan's sleep tracker indicates that she was in REM sleep for 24% of the time last night. Sheridan was in REM sleep for

Nezavi [6.7K]

Answer:

500 minutes

Step-by-step explanation:

Assume you need solution as you have the answer

The full time is x
24% of x is 120 minutes

Find x:

x*24/100 = 120
x *0.24 = 120
x = 120/0.24
x = 500

3 0

3 years ago

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Evaluate b/a - 1.2 when a=4 and b=12

Brums [2.3K]

Answer: 46.8

Step-by-step explanation:

7 0

3 years ago

What’s the least common multiple 5,6,8

mina [271]

Answer:

120

Step-by-step explanation:

List the multiples of 8 and stop when you find a multiple of both 5 and 6. Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, …

multiples of 6 are 6, 12, 18, ....120

multiples of 5 are 5, 10, 15, 20...120

6 0

3 years ago

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Help Determine which equation is parallel to line jk and which is perpendicular to line jk.

alukav5142 [94]

Answer:

5x + 3y = 13 parallel.

6x - 10y = 7 perpendicular.

Step-by-step explanation:

Two lines with slopes $m_1$ and $m_2$ are parallel when $m_1=m_2$ and are perpendicular when $m_1*m_2=-1$

Now determine the slope of all the lines

The line jk passes through the points

(-5,5) and (1,-5) so its slope is

$\frac{-5-5}{1-(-5)}=\frac{-10}{6}=-\frac{5}{3}$

To determine the slope of the lines in the blue rectangles, isolate y from each one and the coefficient of x is the slope

5x-3y = 8 ------> y = (5/3)x + 8/3 ------> slope 5/3

Neither parallel nor perpendicular.

6x+10y = 11 ------> y = (-6/10)x + 11/10 = (-3/5)x + 11/10 ------> slope -3/5

Neither parallel nor perpendicular.

5x + 3y = 13 ------> y = (-5/3)x + 13/3 ------> slope -5/3

This line is parallel

6x - 10y = 7 ------> y = (6/10)x - 7/10 = (3/5)x -7/10 ------> slope 3/5

Since (-5/3)(3/5) = -1 this line is perpendicular.

8 0

3 years ago

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

For the 95% confidence interval we can find the critical value in a t distribution with 50 degrees of freedom and we got:

$t_{\alpha/2}= 2.01$

And replacing we got:

$12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

7 0

4 years ago