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3 years ago

Y=9x-5 in standard form

Mathematics

1 answer:

Mkey [24]3 years ago

3 0

Y equals nine x minus five I’m sorry I don’t know if I did the right thing but yeah

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If the scale factor of a scale drawing is greater than one, the scale drawing is larger than the actual object

galina1969 [7]

False, for example, if the scale factor is 2, the scale drawing would still be smaller because it would be twice as smaller than the actual object.

5 0

3 years ago

There is a quadrilateral MNPQ in which side MN is congruent to side PQ and side NP is parallel to side MQ. The diagonal MP and t

8_murik_8 [283]

Answer: QN = 12

Step-by-step explanation: This quadrilateral is a paralelogram because its 2 opposite sides (NP and MQ) are parallel and the other 2 (MN and PQ) are congruent.

In paralelogram, diagonals bisect each other, which means QR = RN.

If QR = RN:

QR = 6

Then,

QN = QR + RN

QN = 6 + 6

QN = 12

The diagonal QN of quadrilateral MNPQ is QN = 12.

5 0

3 years ago

Which of the following examples illustrates ordinal numbers?

Margaret [11]

Answer:

30-5×2 of 3+(19-3) ÷8

Step-by-step explanation:

30-5×6+(19-3)÷8

30-5×6+16÷8

30-30+8

38-30

6 0

2 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago

-3/2 + 3/7 find the sum. write your answer in simplest form.

DedPeter [7]

Answer: 1 1/14

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers