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3 years ago

The science club has 25 members and 40% of the girls how many girls are in the science club

Mathematics

2 answers:

mario62 [17]3 years ago

8 0

Answer: 10

Step-by-step explanation:25*0.4=10

densk [106]3 years ago

4 0

Answer:

Step-by-step explanation:

25*40/100 = 10

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What is the percent decline, to the nearest whole number, of rusty patched bumble bees from 20 years ago to today?

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Sasuke is pretty cool

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3 years ago

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What is the area of a sector when theta = 15º and r = 4

Afina-wow [57]

Answer:

2.09 sq. units

Step-by-step explanation:

We can simply use the formula for area of a sector of a circle.

Area of sector = $\frac{\theta}{360}*\pi r^2$

Where $\theta$ is the angle and r is the radius.

It is given that the angle is 15 and radius is 4. We plug them in and find the area:

$\frac{\theta}{360}*\pi r^2\\=\frac{15}{360}*\pi (4)^2\\=\frac{1}{24}*16\pi\\=2.09$

Thus area of sector is 2.09 sq. units.

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3 years ago

Which segment is a radius of o 7?

hoa [83]

A is the correct answer

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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

Solve the proportion W over thirty-six = five over eighteen.

storchak [24]

Answer:

$W=10$

Step-by-step explanation:

$\frac{W}{36}=\frac{5}{18}$

multiply both sides by 36

$W=10$

8 0

2 years ago