What is 56.57 rounded to the nearest whole number

Anna and her friends ate pizza for lunch. There were 5 pizzas with 12 slices each. If Anna and her 12 friends had 4 slices each,

Vinil7 [7]

Answer:

Step-by-step explanation:

This is so easy lol what grade are you in

8 0

3 years ago

Read 2 more answers

Jamie needs to build a fence around his garden, as illustrated by polygon ABCDEF on the coordinate grid below. If each unit repr

NNADVOKAT [17]

Answer:

C. 26 yards

Step-by-step explanation:

Jamie's fence total length = perimeter of the polygon

Perimeter of the polygon = AB + BC + CD + DE + EF + FA

AB, FA and DE can be worked accordingly as shown below:

AB = |-5 - 0| = 5 units

FA = |5 - 2| = 3 units

DE = |1 -(-2)| = 3 units

BC, CD, and EF can be calculated using the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between B(0, 5) and C(4, 2):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(0, 5) = (x_1, y_1)$

$C(4, 2) = (x_2, y_2)$

$BC = \sqrt{(4 - 0)^2 + (2 - 5)^2}$

$BC = \sqrt{(4)^2 + (-3)^2}$

$BC = \sqrt{16 + 9} = \sqrt{25}$

$BC = 5 units$

Distance between C(4, 2) and D(1, -2)

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(4, 2) = (x_1, y_1)$

$D(1, -2) = (x_2, y_2)$

$CD = \sqrt{(1 - 4)^2 + (-2 - 2)^2}$

$CD = \sqrt{(-3)^2 + (-4)^2}$

$CD = \sqrt{9 + 16} = \sqrt{25}$

$CD = 5 units$

Distance between E(-2, -2) and F(-5, 2):

$EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$E(-2, -2) = (x_1, y_1)$

$F(-5, 2) = (x_2, y_2)$

$EF = \sqrt{(-5 -(-2))^2 + (2 -(-2))^2}$

$EF = \sqrt{(-3)^2 + (4)^2}$

$EF = \sqrt{9 + 16} = \sqrt{25}$

$EF = 5 units$

Total length of the wall in yards = 5 + 5 + 5 + 3 + 5 + 3 = 26 yards

4 0

3 years ago

4d=1/3 solve you d what is "d"

abruzzese [7]

<span>4d=1/3 [divide both by 4]
d = 1/12

</span>

6 0

3 years ago

Read 2 more answers

Marge and Kimo equally Shared 1/4 of a pie that was left over. What fraction of the original pie did each friend get? Use the pi

luda_lava [24]

Answer:

1/8

Step-by-step explanation:

1/4 divided by 2

8 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.05 in. A one-sample t-test will be carried out

satela [25.4K]

Answer:

a) H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

b) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

c) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

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

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

Step-by-step explanation:

Part a

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part b

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part c

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

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

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

8 0

3 years ago