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Amiraneli [1.4K]

3 years ago

9

Write 39,005 in expanded form

2 answers:

vlada-n [284]3 years ago

8 0

(3 * 10,000) + ( 9* 1,000) + ( 5* 1)

8090 [49]3 years ago

5 0

Just go up by ones tes hundreds thousands and so on to get - 30,000+9000+5

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Multiply: (sqrt10 +2 sqrt8)(sqrt10-2 sqrt8)

vesna_86 [32]

Answer:

(√10 +2√8)(√10 -2√8)=

(10 -8√5 + 8√5 -32)

10+0-32

10-32

= -22

Hope this helps.

6 0

3 years ago

Set up and solve a proportion 6/100=s/12

rusak2 [61]

Amount of tax on the book (s) is $0.72

Step-by-step explanation:

Step 1: Given 6/100 = s/12
Step 2: Cross multiply to solve for s

⇒ 6 × 12 = 100 × s

⇒ 72 = 100s

⇒ s = 72/100 = 0.72

7 0

3 years ago

What is the greatest product you can make by multiplying two 2-digit numbers? Explain.

Anestetic [448]

9801
99 is the greatest 2 digit number so if you do 99x99 you will get 9801

5 0

3 years ago

Read 2 more answers

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

Which point of concurrency is NOT always inside a triangle? Check all that apply. A. circumcenter B. incenter C. orthocenter D.

timama [110]

Answer:

Option A. is correct

Step-by-step explanation:

The circumcenter is a point of intersection of all the perpendicular bisectors of a triangle.

The incenter is a point of intersection of all the angle bisectors of a triangle.

The orthocenter is a point of intersection of all the altitudes of a triangle.

The centroid is a point of intersection of all the medians of a triangle.

The incenter, orthocenter, and centroid always lie inside a triangle.

However, a circumcenter does not always lie inside a triangle.

In an acute-angled triangle, the circumcenter may lie inside or outside the triangle.

So,

Option A. is correct

8 0

3 years ago