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3 years ago

Where to locate the nonnegative abscissa, positive ordinate

1 answer:

4 0

The location to the nonegative abscissa, positive ordinate is in the Q1 it means it is in the fisrt quadrant as seen in the next image: http://www.mathnstuff.com/math/spoken/here/1words/q/q2.htm
Hope this helps

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Simplify each expression. Use a slash () to show a fraction. Do not use any spaces. -(3)^2

I am Lyosha [343]

Answer:

The answer would be -9

Step-by-step explanation:

-(3)^2

-(9)

-9

Following order of operations (PEMDAS)

6 0

2 years ago

Solve 12.01 x 5.21 please help me!

Stells [14]

Answer:

62,5721

Step-by-step explanation:

12,01 × 5,21

= 62,5721

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2 years ago

the ratio of red marbles to blue marbles in a bag is 2 to 3 .which of the following could be the number of red and blue marbles

Andreas93 [3]

Answer:

1/2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Perform the division and leave the result in trigonometric form.

Anettt [7]

Answer:

$\frac{1}{2}$ ( cos40° + isin40°)

Step-by-step explanation:

To divide

$\frac{r_{1(cosx_{1}+isinx_{1}) } }{r_{2}(cosx_{2}+isinx_{2}) }$

= $\frac{r_{1} }{r_{2} }$ [ cos(x₁ - x₂) + isin(x₁ - x₂)

Given

$\frac{2(cos90+isin90}{4(cos50+isin50)}$

= $\frac{2}{4}$ ( cos(90 - 50)° + isin(90 - 50)°

= $\frac{1}{2}$ ( cos40° + isin40° )

8 0

3 years ago

An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0

3 years ago