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stich3 [128]
3 years ago
7

Carman spent 42$ on 6 hats how much did each hat cost. Its for my sisters elementary school class and i dont even know the awnse

r because she hasnt learned division
Mathematics
1 answer:
Travka [436]3 years ago
8 0

Answer: 7

Step-by-step explanation:

Step 1: Since it is asking how much each hat costed. We would divide.

             42$ divide 6 = 7$ per hat


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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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A(2/3x-3)=b solve for x
murzikaleks [220]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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The sum of squares of two consecutive odd numbers is 514 find the numbers ​
amid [387]

Correct Question : The sum of squares of two consecutive positive odd numbers is 514. Find the numbers.

\rule{130}1

Solution :

Let the two consecutive positive odd numbers be x and (x + 2)

\rule{130}1

☯ \underline{\boldsymbol{According\: to \:the\: Question\:now :}}

:\implies\sf x^2 + (x + 2)^{2} = 514 \\\\\\:\implies\sf x^2 + x^2 + 4x + 4 = 514\\\\\\:\implies\sf 2x^2 + 4x + 4 = 514 \\\\\\:\implies\sf 2x^2 + 4x = 514 - 4\\\\\\:\implies\sf 2x^2 + 4x = 510 \\\\\\:\implies\sf 2x^2 + 4x - 510 = 0\\\\\\:\implies\sf x^2 + 2x - 255 = 0 \:\:\:\:\:\Bigg\lgroup \bf{Dividing\:by\:2}\Bigg\rgroup\\\\\\:\implies\sf x^2 + 17x - 15x - 255 = 0\\\\\\:\implies\sf x(x + 17) - 15(x + 17) = 0\\\\\\:\implies\sf (x + 17) (x - 15)\\\\\\:\implies\sf x = - 17\:or\:x = 15\\\\\\:\implies\underline{\boxed {\sf x = 15}}\:\:\:\:\:\Bigg\lgroup \bf{Positive\:odd\: number}\Bigg\rgroup

When,

\bullet\: \textsf {x = } \textbf {15}

\bullet\: \sf x + 2 = 15 + 2 =\textbf{ 17}

\therefore\:\underline{\textsf{The required positive integer is \textbf{15 and 17}}}.

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