One winter day, the temperature ranged from a high of 20 F to a low of -25 F. By how many degrees did the temperature change?

ivanzaharov [21]

The x intercept is 1 and the y intercept is -3

To help you with problems like this, there is a great graphing website called Desmos where you can put in the equation and it will graph it for you. If you need to know how to find the intercepts, I can help with that too.

The x intercept is where the line crosses the x axis (side to side)

The y intercept is where the line crosses the y axis (up and down)

5 0

3 years ago

THIS IS DUE TODAY SO CAN ANYONE PLEASE FIGURE THIS OUT!!

IrinaK [193]

Answer:

I would say B

Step-by-step explanation:

when you are going down you tend to go faster because of gravity. So you would go down, up, down, up, and a little down. This would make B the answer because the speed is increasing when it goes down so that is the only answer that would make sense. Hope this helped! Have a good day! :)

3 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

Please someone help me!!! its urgent!

I am Lyosha [343]

Answer:

12.99

Step-by-step explanation:

a²+b²=c²

a²+b²=c²√91²+√78²=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169 =13

8 0

2 years ago

Find the length of the side of a square whose area is equal to the area of a rectangle with side measures 4.9 m and 1.6 m

trapecia [35]

Given :

Area of square equals to the area of rectangle of sides 4.9m and 1.6m .

To find :-

The side of the square .

Solution :-

As we know ,

Area = side²
4.9 * 1.6 = a²
a = √ 49/10 * 16/10
a = 7/10 * 4/10
a = 28/100
a = 0.28 m

6 0

3 years ago