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2 years ago

How do you solve 3x-15=7x-10

Mathematics

2 answers:

Anna35 [415]2 years ago

8 0

Answer:

I think it’s -1.25???? Idk I’m too lazy to check my math

Step-by-step explanation:

You take the 3x and subtract it from both sides. You now have -15 = 4x -10

Then you do +10 to both sides. Now you have -5 = 4x

Divide each side by 4 and you have -1.25 = x

Now check your math.

3 • -1.25 -15 = 7 • -1.25 -10

Fofino [41]2 years ago

5 0

Answer:

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form: x=-45

−

Decimal Form:x=-1.25

−

1.25

Mixed Number Form:x=-14

−

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A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for

zheka24 [161]

Answer:

The probability is 1/2

Step-by-step explanation:

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

$W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)$

The values of the cummulative distribution of the Standard Normal distribution, lets denote it $\phi$ , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization

$P(X > 50) = P(\frac{X-50}{8.33} > \frac{50-50}{8.33}) = P(W > 0) = \phi(0) = 1/2$

Download pdf

8 0

2 years ago

ACME insurance company estimates that the probability of the owners’ of a house worth between $120,000 to $170,000 making a clai

olchik [2.2K]

The expected value of this policy to the insurance company is $285.00.

Using this formula

Policy expected value=Insurance policy charges-[(Probability × Claim)+(Probability × Claim)]

Let plug in the formula

Policy expected value=$1,300-{(.0041)($150,000)+(.08)($5,000)]

Policy expected value=$1,300-($615+$$400)

Policy expected value=$1,300-$1,015

Policy expected value=$285.00

Inconclusion the expected value of this policy to the insurance company is $285.00

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brainly.com/question/19819099

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2 years ago

What is the domain of the function?<br><br> y= sqrt 5x-10

Andrews [41]

I don't know if this will help but I found this on YAHOO!

Answer: Domain of definition of a function is the set of numbers which the variable attains and for which the function is defined.

Step-by-step explanation:

f(x) = sqrt (5x-10)

Here x can have value equal to any real number >=2 because if x attains value less that 2, 5x-10 becomes negative and sqrt(5x-10) has no real value.

therefore the domain of the function f(x) is (2, infinity) inclusive of 2.

8 0

2 years ago

Find r if 3(r - 7) = 4(2 - 2r) + 4

Arturiano [62]

Answer:

r=3

Step-by-step explanation:

3(r - 7) = 4(2 - 2r) + 4

Distribute

3r -21 = 8 -8r +4

Combine like terms

3r -21 = 12 -8r

Add 8r to each side

3r +8r-21 = 12 -8r+8r

11r -21 = 12

Add 21 to each side

11r -21+21 =12+21

11r = 33

Divide each side by 11

11r/11 = 33/11

r = 3

4 0

3 years ago

Read 2 more answers

One Sunday, 120 days before Christmas, Aldsworth store publishes an advertisement saying ‘120 shopping days until Christmas'. Al

Lena [83]

Answer:

(a)18

(b)1089

(c)Sunday

Step-by-step explanation:

The problem presented is an arithmetic sequence where:

First Sunday, a=1
Common Difference (Every subsequent Sunday), d=7

We want to determine the number of Sundays in the 120 days before Christmas.

(a)In an arithmetic sequence:

$\text{The nth term}, T_n=a+(n-1)d\\T_n \leq 120\\$Therefore:$\\1+7(n-1) \leq 120\\1+7n-7\leq 120\\7n-6\leq 120\\7n\leq 120+6\\7n\leq 126\\$Divide both sides by 7$\\n\leq 18$

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.

Therefore, Aldsworth advertised 18 times.

(b)Next, we want to determine the sum of the first 18 terms of the sequence

1,8,15,...

$\text{Sum of a sequence}, S_n=\frac{n}{2}( 2a+(n-1)d)\\S_{18}=\frac{18}{2}( 2*1+(18-1)*7)\\=9(2+17*7)\\=9(2+119)\\=9*121\\S_{18}=1089$

The sum of the numbers of days published in all the advertisements is 1089.

(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.

6 0

2 years ago