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3 years ago

How do I do number 3

Mathematics

2 answers:

Pavel [41]3 years ago

8 0

The answer would be Either b or c

kogti [31]3 years ago

5 0

you simply have to add all of the numbers together, so it'll be 2+5+12+7+1 or 27, therefore the answer is B

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On a particular day, the wind added 5 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 5 miles

Nata [24]

now, this is pretty much the same as the one before it with Jaime, so I'll do this without much fuss.

recall d = rt.

a = Alfonso's rate

with the wind his speed is a + 5, against it is a - 5, 60 miles with it and 30 miles against it, all in the same time of t hours.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{with the wind}&60&a+5&t\\ \textit{against the wind}&30&a-4&t \end{array} \\\\\\ \begin{cases} 60=(a+5)t\implies \frac{60}{a+5}=\boxed{t}\\\\ 30=(a-4)t\\ -------------\\ 30=(a-4)\left( \boxed{\frac{60}{a+5}} \right) \end{cases} \\\\\\ 30(a+5)=(a-4)60\implies 30a+150=60a-240 \\\\\\ 390=30a\implies \cfrac{390}{30}=a\implies 13=a$

8 0

3 years ago

A shape is made of 12 right triangles of equal size each right triangle has a base of 4 centimeters and a height a 5-cm what is

mart [117]

Answer:

$120\ cm^2$

Step-by-step explanation:

The area $(A_t)$ of a triangle is equal to half the product of its length (L) because of its height (h).

$A_t = 0.5h*L$

The base of the triangles is $L = 4\ cm$

The height of the triangles is $h = 5\ cm$

Then the area of one of the triangles is:

$A_t = 0.5(4)(5)\\A_t = 10\ cm^2$

The figure is formed by 12 equal triangles, then the area of the figure $(A_f)$ is:

$A_f = 12(A_t)\\A_f = 12(10)\\A_f = 120\ cm^2$

3 0

3 years ago

Read 2 more answers

janet is using a recipe for pasta that requires 1 1/2 cups of flour to make 2 servings she is making 6 servings of pasta how muc

Zina [86]

1.5 * 3 = 4.5 cups of flour

4 0

3 years ago

Quadrilateral ABCD has the given properties. What kind of quadrilateral must it be?

Olenka [21]

What are the properties?

6 0

3 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

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

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago