What is the degree measure of arc UTR

Write an equation for the function that includes the following points (2, 32) and (3, 64)

luda_lava [24]

Answer:

y=32x-32

Step-by-step explanation:

this time no algorithm or equation was needed i saw that if the equation was y=32x there would be ordered pairs (1,32) and (2,64) so to delay it by 1 on the x side just subtract the slope from the y - intercept sorry if that doesn't make sense

6 0

3 years ago

Nancy has 62 more erasers than Barbara. Barbara has ferasers. Write the expression that

Anna11 [10]

Answer:

F+63= The amount of erasers Nancy has

4 0

3 years ago

A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms

MissTica

Answer:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

Step-by-step explanation:

For this case we have the following info given:

$\bar X = 39.5$ represent the sample mean

$s =6.6$ represent the sample deviation

$\mu = 30.6$ represent the reference value to test.

$n = 25$ represent the sample size selected

The statistic for this case is given by:

$t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we got:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

3 0

4 years ago

A bear can run up to 18.47 feet per second. How many feet could the bear run in 3.9 seconds? 720.33 feet 7,203.3 feet 72.033 fee

JulsSmile [24]

Answer:

72.033 ft

Step-by-step explanation:

To find this, let us set up a fraction. Since 18.47 = 1 second, y = is 3.9 seconds. Let "y" be the variable.

$\frac{18.47}{1}=\frac{y}{3.9}$