If anyone knows how to do this please help, its for a test and I have to finish the course by 11:59 tonight

Karo-lina-s [1.5K]

150 grams weigh more than 5 ounces

6 0

2 years ago

The area of a rectangle is 65 m2 , and the length of the rectangle is 3 m less than twice the width. find the dimensions of the

salantis [7]

X = <span>width
2x-3 = </span><span>length

</span> $x(2x-3) = 65\\
2x^2 - 3x - 65 = 0 \\ D=b^2-4ac=(-3)^2-4*2*(-65)=529\\
x_{1,2}= \frac{-bб \sqrt{D} }{2a}= \frac{3б 23}{4} \\ x_1=-5 \ \ \O \\ x_2=6.5 \ \$

width = 6.5 m
length = 2x-3 = 2*6.5 - 3 = 10 m

4 0

3 years ago

What is the solution to the equation 1/4x = 5?

puteri [66]

The solution to your equation is:

Solve:
1/4x = 5

Step 1: Multiply both sides by 4
4 * (1/4x) = (4) * (5)

x = 20

Thus your answer is:

Letter Choice (C), x = 20

Hope that helps!!!

5 0

2 years ago

Find the polynomial of least degree to be added to 6x^2+8x^3+17x^2+12x+7 so that 3x^2+4x+1 is a factor. Class ten questions

AURORKA [14]

Hope that really helps you

3 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago