What is the area of this triangle

Mary wants to buy one large pizza, one medium pizza, and three drinks. The drinks cost $p$ dollars each, the medium pizza costs

Korolek [52]

Answer:

$30-8p$

Step-by-step explanation:

Let p represent cost of each drink.

We have been given that the medium pizza costs two times as much as one drink, so the cost of medium pizza would be $2p$ .

We are also told that the large pizza costs three times as much as one drink, so the cost of la pizza would be $3p$ .

We have been given that Mary wants to buy one large pizza, one medium pizza, and three drinks.

Since the cost of one drink is p, so cost of 3 three drinks would be $3p$ .

The total cost of one large pizza, one medium pizza, and three drinks would be $2p+3p+3p=8p$

Mary started with $30, so amount left after all of her purchases would be 30 minus total cost of all purchases.

$30-8p$

Therefore, Mary will have $30-8p$ dollars left after making all of her purchases.

4 0

2 years ago

Jerry weighs 95 pounds. This is 15 pounds less than Mikey weighs. How much does Mikey weigh? *

monitta

Answer:

Mikey weighs 110 pounds.

Step-by-step explanation:

Because Mikey weighs 15 pounds more, just add 95 pounds and 15 pounds to get Mikey's weight:

95 + 15 = 110

I hope this helps you!

Please rate my answer and give me Brainliest!

Have a nice day! :)

6 0

3 years ago

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Reduce 720/1080 little cunfused

Veronika [31]

To reduce a fraction, divide the numerator and the denominator equally until they reach the simplest whole number possible.

In this case, the numerator (720) and the denominator (1080) can both be divided by 360 to get 2/3, our reduced fraction.

8 0

3 years ago

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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

3 years ago

Solve this root2x-7 =7

Karo-lina-s [1.5K]

-14??
Idek tbh it’s kinda hard

5 0

2 years ago

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