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3 years ago

15

Plsss help this is a math problem

2 answers:

Darina [25.2K]3 years ago

5 0

There are 4 terms in the expression

Nostrana [21]3 years ago

4 0

I think 4 I might not be correct though!

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Boone fitnes sells gym passes in packs of 14. A competing gym, sells gym passes in sets of 20. If both sold the same number of p

sergeinik [125]

20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

14: 14, 28, 42, 56, 70, 84, 98, 112, 126, 140

140 is the smallest number they have in common.

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4 years ago

Subtract 8 feet from 9 yards, 1 foot.

Nostrana [21]

Answer: B

Step-by-step explanation:

I knew I would have 1 foot left so that made sure it wasn’t A then I knew that it the answer had to be in yards so I convert 8 feet into yards which becomes 2.7 so it’s not three which makes I’m not C. So 9 yards minus 2.7 yards would equal the answer B

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3 years ago

Read 2 more answers

The average wavelength in a series of ocean wave is 16.0m. A wave crest arrives at the shore on average every 10.0, so the frequ

lesya692 [45]

Answer:

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wunuwu unwind and relax and enjoy the rest of the rest of the rest

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3 years ago

What is the experamental probability that the coin lands on heads?

sammy [17]

Answer:

The experamental probability that the coin lands on head is 50 %

Step-by-step explanation:

Given:

Experiment:

A coin is Toss

Let the Sample Space be 'S' that is total number of outcomes for a coin has been tossed = { Head, Tail }

∴ n ( S ) = 2

Let A be the event of getting a Head on tossing a coin i.e { Head }

∴ n( A ) = 1

Now,

$\textrm{Probability} =\frac{\textrm{outcomes for the experiment}}{\textrm{total number of outcomes}}$

Substituting the values we get

$P(A) = \frac{n(A)}{n(S)} \\\\P(A) = \frac{1}{2} \\\\P(A) = 0.5\\\\P(A) = 50\%$

The experamental probability that the coin lands on head is 50 %

5 0

3 years ago

The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Artemon [7]

Answer:

P(X $\geq$ 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N( $\mu=73,\sigma^{2}=3^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 73

$\sigma$ = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X $\geq$ 74)

P(X $\geq$ 74) = P( $\frac{X-\mu}{\sigma}$ $\geq$ $\frac{74-73}{3}$ ) = P(Z $\geq$ 0.33) = 1 - P(Z < 0.33)

= 1 - 0.62930 = 0.3707

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

3 0

3 years ago