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3 years ago

You move left 1 unit and up 1 units, you end at (4,-2) where did u start?

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

6 0

Answer:

can you show the like board thing

Step-by-step explanation:

DIA [1.3K]3 years ago

6 0

Answer:

(3,-1)

Step-by-step explanation:

If you look at a graph and plot the point (4,-2) and move left 1 unit, you end up at 3. Then you move up 1 unit. Now you are at -1. Therefore, your point is (3,-1).

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In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position wh

Lady_Fox [76]

Answer: = ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain x = 0.41

Standard deviation r = 0.016

Number of samples n = 1000

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

0.41+/-1.96(0.016/√1000)

0.41+/-1.96(0.000506)

0.41+/-0.00099

0.41+/-0.001

= ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

4 0

4 years ago

Which graph shows the line y = x + 1?

Iteru [2.4K]

I need a picture in order to help

Here is a picture of a bobcat anyway

6 0

3 years ago

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The greatest common factor of 3m2n + 12mn2 is

Kamila [148]

$3m^2n+12mn^2$
In both the sets, 3 and m as well as n can be found as common factors.
So,
3mn( m+4n)

Therefore the answer is 3mn (#3)

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3 years ago

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Plz help! What is the volume of the composite figure?!

maria [59]

F f f. Fbtbtbt bzubshbsbydby

3 0

3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago