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2 years ago

−3x^2−4x+7 answer this problem please - algebra - factoring

Mathematics

2 answers:

Airida [17]2 years ago

8 0

Answer:

-(x-1) (3x+7)

Step-by-step explanation:

disa [49]2 years ago

7 0

Answer:

-((x - 1)(3x + 7))

Step-by-step explanation:

You might be interested in

Can someone please help me factor this

Dmitry_Shevchenko [17]

Answer:

$\huge\boxed{\bf\:1}$

Step-by-step explanation:

$\frac{ x ^ { 2 } -4x+3 }{ x ^ { 2 } -7x+12 } \times \frac{ x ^ { 2 } +2x-24 }{ x ^ { 2 } +5x-6 } ^ { }$

Take $\frac{ x ^ { 2 } -4x+3 }{ x ^ { 2 } -7x+12 }$ & factorise it at first.

$\frac{ x ^ { 2 } -4x+3 }{ x ^ { 2 } -7x+12 } \\= \frac{\left(x-3\right)\left(x-1\right)}{\left(x-4\right)\left(x-3\right)}\\= \frac{x-1}{x-4}$

Now factorise the next set : $\frac{ x ^ { 2 } +2x-24 }{ x ^ { 2 } +5x-6 } ^ { }$ .

$\frac{ x ^ { 2 } +2x-24 }{ x ^ { 2 } +5x-6 } ^ { }\\= \frac{\left(x-4\right)\left(x+6\right)}{\left(x-1\right)\left(x+6\right)}\\= \frac{x-4}{x-1}$

Now, multiply the two simplified results.

$\frac{ x ^ { 2 } -4x+3 }{ x ^ { 2 } -7x+12 } \times \frac{ x ^ { 2 } +2x-24 }{ x ^ { 2 } +5x-6 } ^ { }\\= \frac{x-1}{x-4}\times \frac{x-4}{x-1} \\= \frac{\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-1\right)} \\= \boxed{\bf\: 1}$

$\rule{150pt}{2pt}$

7 0

2 years ago

The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab

kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(d) 240

Step-by-step explanation:

The random variable X can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable X thus follows a Poisson distribution with parameter, λ = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.

The mean of the approximated distribution of X is:

μ = λ = 250

The standard deviation of the approximated distribution of X is:

σ = √λ = √250 = 15.8114

Thus, $X\sim N(250, 250)$

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

$P(235$

$=P(-0.95$

Thus, the value of P (235 < X < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.28$

Thus, the value of $P(48$ .

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.40$

Thus, the value of $P(48$ .

(d)

Let the sample size be n.

$P(48$

$0.95=P(-z$

The value of z for this probability is,

z = 1.96

Compute the value of n as follows:

$z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241$

Thus, the sample selected must be of size 240.

5 0

3 years ago

The table shows the number of cups of water required when cooking different amounts of rice. Amount of Rice (cups) Amount of Wat

Ugo [173]

Amount of Rice (cups) Amount of Water (cups)
2 5
3 7.5
5 12.5
8 20

Let's find the ratio of rice and water:

5 / 2 = 2.5 => 2 * 5 = 5

7.5 / 3 = 2.5 => 3 * 2.5 = 7

12.5 / 5 = 2.5 => 5 * 2.5 = 12.5

20 / 8 = 2.5 => 8 * 2.5 = 20

So, as the ratio of water ot rice is constant, and the first variable is the amount of rice, you can calculate the amount of water given the amount of rice.

So, the amount of rice is independent value and the amount of water is dependent (it depends on the amount of rice).

The amount of rice is the dependent value. FALSE
The amount of water is the dependent value.TRUE
The amount of rice is the independent value. TRUE
The amount of water is the independent value.FALSE

7 0

3 years ago

Read 2 more answers

A television weighs 7.4 kilograms how many grams does the television weigh

Pavlova-9 [17]

7400 grams is the answer.

6 0

3 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

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 high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

2 years ago