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3 years ago

What method most efficient to use to solve x^2+4x+3=0

1 answer:

6 0

My recommendation would be to factor..
x² + 4x + 3 = 0 factor the left side into two binomials
(x + 3)(x + 1) = 0 set each of these binomials equal to zero and solve

x + 3 = 0 Implies x = -3
x + 1 = 0 implies x = -1

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A jet plane is traveling at 840 km/h at 37,000 ft between NY and SF. What equation shows the relationship between total kilomete

fgiga [73]

What equation shows the relationship between total kilometers traveled (m), time spent traveling (t), and the plane's speed (s)?

m = t * s

7 0

3 years ago

Rewrite the formula to find w.

sveta [45]

Answer:

P / 30 - 21= 9

2w = 9

w = 4.5

This should be the answer

4 0

3 years ago

To play at bingo night there’s a $10 cover charge +2 dollar per bingo card. Write an equation for the cost (y) based on the numb

jasenka [17]

Answer:

y = 2x+10

Step-by-step explanation:

y is the cost (which we don't know)

2x is the $2 per card with x representing the number of cards (which we don't know)

And of course 10 is the cover charge.

4 0

3 years ago

The length of a rectangle is 6 in longer than its width.if the perimeter of the rectangle is 68in find the length and width ?

wel

The length and the width of the rectangle are 20 and 14 inches respectively

<h3>How to find the length and width?</h3>

The given parameters are:

Length = 6 + Width

Perimeter = 68 inches

The perimeter of a rectangle is calculated as:

Perimeter = 2 * (Length + Width)

So, we have:

2 * (Length + Width) = 68

Divide both sides by 2

Length + Width = 34

Substitute Length = 6 + Width in Length + Width = 34

6 + Width + Width = 34

Evaluate the like terms

2 * Width = 28

Divide both sides by 2

Width = 14

Substitute Width = 14 in Length = 6 + Width

Length = 6 + 14

Evaluate

Length = 20

Hence, the length and the width of the rectangle are 20 and 14 inches respectively

Read more about perimeter at:

brainly.com/question/24571594

#SPJ1

6 0

1 year ago

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin

Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let $\mu$ = true average writing lifetime under controlled conditions

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

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

where, $\bar X$ = sample mean

s = sample standard deviation

n = sample size of pens = 18

n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then we will reject our null hypothesis as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0

3 years ago