Do 3/7and7/3 represent the same length on a number line?

Fantom [35]

Answer:

the area of the hexagon is approx. 187.1 in²

Step-by-step explanation:

Picture this regular polygon as being a hexagon made up of six equilateral triangles of side 12 in. We find the area of one such triangle and then multiply that by 6 to obtain the total area of the hexagon.

One such equilateral triangle has three sides all of length 12 in, and all the interior angles are 60°. The height of one such triangle is

h = (12 in)sin 60°, or

√3

h = (12 in) -------- = 6√3 in

2

So, with base 12 in and height 6√3 in, the area of one such equilateral triangle is

A = (1/2)(12 in)(6√3 in) = 36√3 in²

and the total area of the hexagon is 6(36)√3 in², or approx. 187.1 in²

5 0

3 years ago

Algebra 1 Check Answer Please

BigorU [14]

The answer is the third one.

The way I think through these problems is that where you multiply (-7x), you use addition, and where you use addition (-4), you multiply.

The answers added together will equal the second number, -7, and multiplied will give you the third, -4.

What you had chosen as the answer was close. But 7.53+-0.53 equals positive 7, not negative. However they do round to -4 when multiplied.

The correct answer, -7.53 and 0.53, do equal -7 when added together and equal -4 when multiplied.

5 0

3 years ago

A bathtub that holds 22 gallons of water contains 12 gallons of water. You begin filling it, and after 5 minutes, the tub is ful

Sedaia [141]

Answer:

8 Gallons Per Minute

Step-by-step explanation:

7 0

2 years ago

mrs.jones is thinking of a number thst when rounded to nearest hundred is 400 what is the number explain

kherson [118]

The number could be several different ones. Are you sure that's exactly what the problem says? If not, I'm going to ask you write the entire problem. If not, then whoever made it did not give enough information for there to be an absolute answer. I'm sorry I couldn't help you, but I really hope I can with more information.

6 0

3 years ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

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

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

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

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

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

3 years ago