A youth Basketball has a standard Circumference of 27 inches. An NBA standard ball is 29.5 inches.

$72 with a 40% markup

alex41 [277]

Answer:

$100.80

Step-by-step explanation:

40%/100%×72=$28.80

$72+$28.80=$100.80

8 0

3 years ago

Suppose you choose a marble from a bag containing 5 red marbles, 3 white marbles, and 6 blue marbles. You return the first marbl

Aneli [31]

$|\Omega|=14^2=196\\
|A|=5\cdot6\cdot2=60\\\\
P(A)=\dfrac{60}{196}=\dfrac{15}{49}$

5 0

3 years ago

Find the area of the shaded region to the nearest square unit. The base and height of the smaller shaded right

Vikentia [17]

Answer:

d

Step-by-step explanation:

let side of smaller triangle=x

x²+x²=7²

2x²=49

$x=\frac{7}{\sqrt{2}} \\area ~of~smaller~triangle=\frac{1}{2} *\frac{7}{\sqrt{2}} *\frac{7}{\sqrt{2}} =\frac{49}{4}\\Total~area=\frac{1}{2} *14*14+12.25 =98+12.25=110.25 ~sq.~units$

=110 sq. ft

4 0

3 years ago

Pls pls pls help if ur good with angles and sides

liq [111]

Answer:

∆PQR: Equilateral Triangle

∆PRT: Scalene Triangle

∆TQS: Isosceles Triangle

∆QNP: Right Triangle

Step-by-step explanation:

∆PQR is an Equilateral Triangle because all of its angles are equal (60°).

∆PRT is a Scalene Triangle because none of its sides are equal.

The triangle above it is equilateral, and one side length is 14, meaning all of its side lengths are 14.

This makes the top of ∆PRT 14. The hypotenuse is given and it’s 18.

The last side left is clearly shorter than both, meaning none of the sides are equal.

∆TQS is an Isosceles Triangle because 2 angles are equal.

The base angles are both 76°.

∆QNP is a Right Triangle because segment QN and PR are perpendicular, creating two 90° angles on both sides.

For number 10, I suggest using a graphing calculator or desmos to graph the points. (Or you can just add the points but I suggest graphing).

From there, you can count the length of each side.

Segment DE: 9

Segment EF: 8

Segment FD: 10

This makes ∆DEF in number 10 a Scalene Triangle because none of its sides are equal.

For number 11, I also suggest graphing the points and counting the sides.

Segment DE: 7

Segment EF: 9

Segment FD: 7

This makes ∆DEF in number 11 an Isosceles Triangle because 2 of its sides are equal.

Hope this helps! :)

4 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago