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3 years ago

One might be as small as 0.000000314 meter in length.how is this length expressed in scientific notation?

Mathematics

1 answer:

devlian [24]3 years ago

5 0

Answer:

3.14 x 10^-7

Step-by-step explanation:

move the decimal behind the first significant figure and count the amount of spaces from there to the original decimal place

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How do I solve this ?

eduard

Look at the picture.

ΔZYX and ΔDYC are similar. Therefore the lengths of sides are in proportion:

$\dfrac{DC}{ZX}=\dfrac{DY}{ZY}$

DC = x

ZX = 16

DY = a

ZY = a + a = 2a

Substitute:

$\dfrac{x}{16}=\dfrac{a}{2a}$

$\dfrac{x}{16}=\dfrac{1}{2}$ multiply both sides by 16

$\boxed{x=8}$

<h3>Answer: CD = 8</h3>

6 0

3 years ago

First.

velikii [3]

Answer:o

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the degree of the polynomial and what the y-intercept

Arada [10]

Answer:

Polynom degree: 5

Y intercept point: (0, 80)

Step-by-step explanation:

P(x)=(x+5)(x+4)²(x+1)²

When you expand, the highest power of x is 1 for first term (x+5), 2 for second term (x+4)² and again 2 for (x+1)². Overall, x⁵ will be the x term with highest power. So the degree of the polynom is 5.

The y intercept, i.e. intersection with OY axis, happens for x=0. Thus, P(0)=5×4²×1²=5×16=80. The y intercept point is (0, 80)

5 0

3 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0

3 years ago

5. What’s the answer to this question ASAP

tatiyna

Answer:

D is the correct answer

(hope this helps)

8 0

3 years ago