HEY *25 POINTS* HELP WITH MY ANGLES PLEASE PLEASE

Toys R Us is having a 38 percent discount on all of their backpacks if a back pack is normally 22.00 how much will it be on sale

My name is Ann [436]

Answer:

toys r us is shut down but to answer the price would be $13.64 saving you $8.36 have a wonderful day

Step-by-step explanation:

6 0

3 years ago

A prop assistant needs to rent a telescope to use for a scene in a movie. One company charges $57 per day plus a flat fee of $32

vovangra [49]

The number of days for both Companies to have the same total amount is

6.4 days

<h3>Word Problem leading to Algebraic expression</h3>

Given Data

Let the total payment be y

Let the number of days be x

Company A

Charges per day = $57

Flat fee = $32

y = 57x+ 32 -----------------1

Company B

Charges per day = $62

Flat fee = $0

y = 62x + 0-----------------2

Equating 1 and 2 we have

57x+ 32 = 62x+ 0

Solving for x we have

62x-57x = 32

5x = 32

Divide both sides by 5

x = 32/5

x = 6.4 days

Learn more about word problem here:

brainly.com/question/13818690

5 0

3 years ago

Quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm. Which could be the side lengths of a dilation of HIJK wit

matrenka [14]

Answer:

(C) 18cm, 39cm, 21cm and 45cm.

Step-by-step explanation:

The quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm.

When HIJK is dilated with a scale factor of 1.5, the side lengths becomes:

12 X 1.5 =18 cm

26 X 1.5 =39 cm

14 X 1.5 =21 cm

30 X 1.5 =45 cm

A dilation of HIJK with a scale factor of 1.5 will give us the side lengths:

18cm, 39cm, 21cm and 45cm.

<u>The correct option is C.</u>

6 0

3 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

4 years ago

I never got my anwser to finding the product and write the lowest term to<br> 7/54×27/35

kenny6666 [7]

Answer:

Step-by-step explanation:

7 / 54 * 27/35 =

1/2 * 1/5 =

1/10 <===

4 0

4 years ago