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4 years ago

1. Factor x²+5x-24 2. Factor 2n²-5n-3

Mathematics

1 answer:

Fynjy0 [20]4 years ago

6 0

Given problems;

1. Factor; x²+5x-24

To factor this equation, we must understand that this is a quadratic equation.

A quadratic equation usually has two roots.

Equate to zero;

x²+5x-24 = 0

Then find two numbers whose product is -24 and their sum is +5

the numbers are +8 and -3

then replace 5x with the number;

x² + 8x - 3x - 24 = 0

Now factorize;

x(x +8) - 3(x + 8) = 0

(x-3)(x+8) = 0

Factoring gives (x-3)(x+8)

2. Factor 2n² - 5n -3;

Equate to zero;

2n² - 5n -3 = 0

Find the number whose sum is -5 and the product gives -6( 2 x -3);

the number is -6 and +1;

2n² - 6n + n -3 = 0

2n(n-3) + 1(n-3) = 0

(2n + 1)(n-3) = 0

The factor of the equation is (2n + 1)(n-3).

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Answer: D

Step-by-step explanation:

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3 years ago

What is the slope of the line that passes through the pair of points (2, 5) and (8, 3)

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Answer:

slope = - $\frac{1}{3}$

Step-by-step explanation:

Calculate the slope m using the slope formula

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4 years ago

A sea turtle swims at a speed of 27 kilometers per hour. A girl swims 14 decimeters per second. 1 m = 10 dm 1000 m = 1 km How mu

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The sea turtle's speed is
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4 years ago

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Each person in random samples of 207 male and 253 female working adults living in a certain town in Canada was asked how long, i

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Answer:

The null hypothesis failed to be rejected.

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Step-by-step explanation:

The data is:

Males n1=207 M1=31.6 s1=24.0

Females n2=253 M2=29.3 s2=24.3

This is a hypothesis test for the difference between populations means.

The claim is that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

Then, the null and alternative hypothesis are:

$H_0: \mu_m-\mu_f=0\\\\H_a:\mu_m-\mu_f> 0$

The significance level is 0.05.

The sample 1 (males), of size n1=207 has a mean of 31.6 and a standard deviation of 24.

The sample 2 (females), of size n2=253 has a mean of 29.3 and a standard deviation of 24.3.

The difference between sample means is Md=2.3.

$M_d=M_m-M_f=31.6-29.3=2.3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{24^2}{207}+\dfrac{24.3^2}{253}}\\\\\\s_{M_d}=\sqrt{2.783+2.334}=\sqrt{5.117}=2.262$

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The degrees of freedom for this test are:

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This test is a right-tailed test, with 458 degrees of freedom and t=1.017, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.017)=0.155$

As the P-value (0.155) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

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3 years ago