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3 years ago

If 5 times a number plus two is 17 what is the number

Mathematics

2 answers:

sweet [91]3 years ago

8 0

Answer:

Step-by-step explanation:

by question

5x+2=17

5×=15

×=3

Fantom [35]3 years ago

5 0

Answer:

The number is 3.

Step-by-step explanation:

5 x 3 = 15

15 + 2 = 17

You can figure this out if you subtract 2 from 17, and then divide that number (10) by 5.

Hope this helped! :)

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15 POINTS-

Doss [256]

I think it’s $3.55
I think this because if u subtract 2.45- 45% = 1.10 sooo u will have to add $1.10 to 2.25 and that will add up the 45% that Walmart is paying with the original price.

8 0

2 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago

Please answer Who is wrong and why

Marysya12 [62]

Answer:

Hi I have to tell you But I also use Flvs Open lesson 01.07 and read it I promise you will find something of how to do it because I had the same prob and after reading the lesson and awnsering the quetion I GOT A 100 hope i help you

Step-by-step explanation:

6 0

2 years ago

The U.S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is regi

yan [13]

Answer:

We conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

Step-by-step explanation:

We are given that 513 employed persons and 604 unemployed persons are independently and randomly selected, and that 287 of the employed persons and 280 of the unemployed persons have registered to vote.

Let $p_1$ = percentage of employed workers who have registered to vote.

$p_2$ = percentage of unemployed workers who have registered to vote.

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$ {means that the percentage of employed workers who have registered to vote does not exceeds the percentage of unemployed workers who have registered to vote}

Alternate Hypothesis, $H_A$ : $p_1>p_2$ {means that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of employed workers who have registered to vote = $\frac{287}{513}$ = 0.56

$\hat p_2$ = sample proportion of unemployed workers who have registered to vote = $\frac{280}{604}$ = 0.46

$n_1$ = sample of employed persons = 513

$n_2$ = sample of unemployed persons = 604

So, the test statistics = $\frac{(0.56-0.46)-(0)}{\sqrt{\frac{0.56(1-0.56)}{513}+\frac{0.46(1-0.46)}{604} } }$

= 3.349

The value of z test statistics is 3.349.

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistic is more than the critical value of z as 3.349 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

5 0

2 years ago

Make a frequency distribution and find the relative frequencies for the following number set. Round the relative frequency to th

Mandarinka [93]

Number......frequency.....relative frequency
10 2 2/24 = 8.3%
20 2 2/24 = 8.3%
30 4 4/24 = 16.7%
40 2 2/24 = 8.3%
50 3 3/24 = 12.5%
60 5 5/24 = 20.8%
70 3 3/24 = 12.5%
80 1 1/24 = 4.2%
90 2 2/24 = 8.3%
total : 24

7 0

3 years ago

If 5 times a number plus two is 17 what is the number​

If 5 times a number plus two is 17 what is the number