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2 years ago

8

The Japanese 5 yen coin has a hole in the middle. The diameter of the coin is 22 mm and the diameter of the hole is 5 mm. What i

s the difference between the outer edge and the inner edge of the coin? If necessary, round your answer to the nearest tenth.

2 answers:

ololo11 [35]2 years ago

8 0

Answer:20mm

Step-by-step explanation:

Lorico [155]2 years ago

7 0

Answer:

Sample response: To find the difference between the outer and inner edges of the coin, you would calculate the circumference of the coin, C= 22π, and the circumference of the hole, C = 5π. Then you would subtract to find the answer, 22π − 5π = 17π.

<em>hope you have an awesome day/night :)</em>

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Of the 24,000 cellular phones inspected in the last production run , 50 were defective . What fraction of the cellular phones in

Nina [5.8K]

Answer:

Step-by-step explanation:

24,000 inspected.....50 were defective

50/24,000 reduces to 1/480 <===

4 0

3 years ago

In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

marissa [1.9K]

Answer:

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

$YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X$ (1)

If we know that $X = 49^{\circ}$ , $XY = 18$ and $YZ = 14$ , then we have the following second order polynomial:

$14^{2} = XZ^{2} + 18^{2} - 2\cdot (18)\cdot XZ\cdot \cos 49^{\circ}$

$XZ^{2}-23.618\cdot XZ +128 = 0$ (2)

By the Quadratic Formula we have the following result:

$XZ \approx 15.193\,\lor\,XZ \approx 8.424$

There are two possible triangles, we can determine the value of angle Y for each by the Law of Cosine again:

$XZ^{2} = XY^{2} + YZ^{2} - 2\cdot XY \cdot YZ \cdot \cos Y$

$\cos Y = \frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ}$

$Y = \cos ^{-1}\left(\frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ} \right)$

1) $XZ \approx 15.193$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 54.987^{\circ}$

2) $XZ \approx 8.424$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 27.008^{\circ}$

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

6 0

3 years ago

Starting in the 1970s, medical technology allowed babies with very low birth weight (VLBW, less than 1500 grams, about 3.3 pound

cluponka [151]

Answer:

The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

The significance level is 0.05.

The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.

$p_1=X_1/n_1=188/244=0.77049$

The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.

$p_2=X_2/n_2=197/247=0.79757$

The difference between proportions is (p1-p2)=-0.02708.

$p_d=p_1-p_2=0.77049-0.79757=-0.02708$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{188+197}{244+247}=\dfrac{385}{491}=0.988$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.988*0.012}{244}+\dfrac{0.988*0.012}{247}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00005}=\sqrt{0.0001}=0.0098$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.02708-0}{0.0098}=\dfrac{-0.02708}{0.0098}=-2.76$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(t$

As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

3 0

3 years ago

What is the primary goal and secondary goal of the Fried Liver Opening in chess? <br />

polet [3.4K]

Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.

Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.

3 0

3 years ago

Never mind no one will answer it

Vinil7 [7]

What
ask the question I will answer it
go ahead ask

5 0

3 years ago

Read 2 more answers