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3 years ago

a rectangle has length 3 cm greater than its width.it has an area of 28cm squared,find the dimensions of the rectangle

1 answer:

3 0

Let w be the width
And let h be the length
h = w + 3

w * h = 28
w(w+3) = 28
w^2 + 3w = 28
Use a calculator (or use the quadratic formula) to get w=4

To find h, just add w by 3.
You get 7.

The dimensions of the rectangle is 4 by 7 cm.

Have an awesome day! :)

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If sin x = 12÷13 where 0° <x<90°<br>find the value of 1-cos ×cos( x)

s2008m [1.1K]

Answer:

This is how tackle this type of problem:

Draw a right triangle with the indicated sides: side opposite the angle = 12 and hypotenuse = 13

Then, calculate the 3rd side by Pythagoras: x2 + 122 = 132

x2 = 169 - 144

x2 = 25

x = 5

Now, we have all 3 sides of our diagram triangle, so we can find all the trig ratios for the angle.

Next, we use the double angle formula to simplify cos(x + 30) = cos x cos 30 - sin x sin 30

= 5/13 . √3/2 - 12/13 . 1/2

= 5√3 /26 - 6/13

= (5√3 - 12) / 26

Step-by-step explanation:

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3 years ago

What is the length of a 50-inch TV that has a width of 27 inches?

drek231 [11]

Answer:

42.08inches

Step-by-step explanation:

Given data

SIze of TV= 50 inch

Width= 27inches

Length= ??

Applying Pythagoras theorem

z^2= x^2+y^2

50^2= x^2+27^2

square root both sides

50^2= x^2+27^2

2500= x^2+ 729

2500-729=x^2

1771=x^2

x= √1771

x= 42.08

Hence the lenght is 42.08inches

7 0

3 years ago

Evaluate the discriminant.<br> Exact answer (no rounding)<br> 0= -4x – 24x – 38

frutty [35]

Answer: x= -19/14 —> (19 over 14)

8 0

3 years ago

A desk manufacturer claims that the average time it takes to assemble a desk is 90 minutes with a standard deviation of 32 minut

stealth61 [152]

Answer:

The standard deviation of the sample mean is 4 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

The standard deviation of the population is 32.

Sample of 64.

$s = \frac{32}{\sqrt{64}} = 4$

The standard deviation of the sample mean is 4 minutes

6 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago