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2 years ago

Please look below for the question :)

Mathematics

2 answers:

Serjik [45]2 years ago

8 0

Answer:

$59.50

Step-by-step explanation:

Lets work out the equation for the given relationship.

This is a linear relationship.

Find the slope first:

(21.75 - 19.50)/(3 - 2) = 2.25

Use one pair of points to get the equation:

c - 19.50 = 2.25(m - 2)
c - 19.50 = 2.25m - 4.50
c = 2.25m + 15

We see the yearly fee is $15 and monthly rate is $2.25.

If we apply the changes to rates the equation becomes:

c = (2.25*2)m + (15 - 5) = 4.5m + 10

Find the cost of 11 months of membership:

c = 11*4.5 + 10 = $59.50

rewona [7]2 years ago

5 0

Answer:

well i think its over 103.10 because 11 months divide 2 fraction then take 1 carry or instead use calculator and type 0077239812 this binary!! and divide by 23 and thats the answer its only like decimals from the point!!

Step-by-step explanation:

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Find the variable of x. 8x-2=-9+7x

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Answer:

8x-2=-9+7x equals X=-7

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3 years ago

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Given that an adult likes soccer, what is the probability the adult is aged 18–30? 8.1% 18.1% 24.5% 44.4%

dsp73

The probability that an adult likes soccer is aged between 18–30 will be 44.4%.

We have an adult who likes soccer.

We have to determine the probability the adult is aged 18–30.

<h3>What is Probability?</h3>

The formula to calculate the probability of occurrence of an event 'A' can be written as -

P(A) = $\frac{n(A)}{n(S)}$

where -

n(A) = Number of outcomes favorable to event A.

n(S) = Total number of outcomes.

According to question, we have an adult likes soccer.

The answer to this question is based on the hypothesis that the adults between 18 - 30 are highly energetic. To be more precisely - the adults in the range 18 - 24 and 24 - 30 are highly energetic and full of stamina. Above the age 30, the number of adults who like soccer will start to decrease and will hit nearly zero between the age range of 55 - 65 as the adults in this age group found it very difficult to even walk.

Mathematically -

The probability of an event A = an adult likes soccer is aged between 18–30 will be the highest value among the ones mentioned in options.

Hence, the probability that an adult likes soccer is aged between 18–30 will be 44.4%.

To solve more questions on Probability, visit the link below -

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1 year ago

To date, this year, Company XYZ has sold 450 units. They sell units at an average rate of 15 per week. The company wants to sell

Arisa [49]

The inequality used to solve for x is x=750/15

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3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

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3 years ago

Can anyone help me with my math i'm really behind on it.

statuscvo [17]

Answer:

yes

Step-by-step explanation:

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