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2 years ago

What is theperature of this object

Mathematics

1 answer:

Andrei [34K]2 years ago

6 0

Answer:

59-60 F and 15 C

Step-by-step explanation:

Hope this helps!

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I need help solving this please?

docker41 [41]

The correct answer is C

4 0

3 years ago

17. Factor x ^ 3 + 5x ^ 2 - 9x - 45 . A. (x - 5)(x ^ 2 + 9) b . (x - 5)(x - 3) * (x + 3) Factor By Groupin (x + 5)(x - 3) * (x +

jok3333 [9.3K]

Answer:

Step-by-step explanation:

Given

x³ + 5x² - 9x - 45 ( factor the first/second and third/fourth terms )

= x²(x + 5) - 9(x + 5) ← factor out (x + 5) from each term

= (x + 5)(x² - 9) ← factor as a difference of squares

= (x + 5)(x - 3)(x + 3)

3 0

3 years ago

Alex drew a scale drawing of an auditorium. He used the scale 3 inches = 2 feet. If the stage is 30 inches in the drawing, how w

White raven [17]

Answer:

20 feet

Step-by-step explanation:

The scale is 3 inches is to 2 feet

As a ratio, it would be:

$\dfrac{3}{2}$

The question is what is the actual length of the stage if the drawing is 30 inches. All we need to do is find a ratio proportional to the scale ratio:

$\dfrac{3inches}{2feet} = \dfrac{30inches}{xfeet}$

Now to find the missing number, cross multiply the fractions first:

$\dfrac{3}{2}=\dfrac{30}{x}$

$(3)(x)=(2)(30)$

$3x=60$

Then solve for x:

$3x=60$

$x=\dfrac{60}{3}$

$x=20$

The answer is 20 feet

8 0

3 years ago

35. In a simple random sample of 25 high school students, the sample mean of the SAT scores was 1450, and the sample variance wa

Tatiana [17]

Answer:

$1450-2.0\frac{30}{\sqrt{25}}=1450-12$

$1450+2.0\frac{30}{\sqrt{25}}=1450+12$

And the best option would be:

c. 1450 +/- 12

Step-by-step explanation:

Information provided

$\bar X=1450$ represent the sample mean for the SAT scores

$\mu$ population mean (variable of interest)

$s^2 = 900$ represent the sample variance given

n=25 represent the sample size

Solution

The confidence interval for the true mean is given by :

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

The sample deviation would be $s=\sqrt{900}= 30$

The degrees of freedom are given by:

$df=n-1=2-25=24$

The Confidence is 0.954 or 95.4%, the value of $\alpha=0.046$ and $\alpha/2 =0.023$ , assuming that we can use the normal distribution in order to find the quantile the critical value would be $z_{\alpha/2} \approx 2.0$