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3 years ago

What is 0.316 in expanded form.

2 answers:

7 0

Uhm, I'm pretty new to expanded form but I think it's something like:
(3 × 1/10) + (1 × 1/100) + (6 × 1/1000)
You can also write it as:
0.3 + 0.01 + 0.006
Let me know if you need working or anything!

IRINA_888 [86]3 years ago

5 0

0.3 + 0.01 + 0.006. ................

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JUST 1 PLEASEEEEEEEEEE AND HOW DO YOU KNOW!!!!! PLEASEEEEEEEEEEEEE HELLLPPPPPPPPPPPP MEEEEEEEEEEEEEE

Lemur [1.5K]

Answer:

I can't see three but I'll answer the first two.

Step-by-step explanation:

Number one: The first, third, and fifth options are correct,

Number two: A is correct because if you divide 3/4 you get 0.75. Check by multiplying 0.75 for one by 4 and you get three dollars.

So for three, one pound of pears costs two dollars. That is consistent with the graphs. Pick the answer that states that the best.

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2 years ago

20) Laila borrowed $9000 from a bank that

iris [78.8K]

Answer:

the answer i think is $259200

Step-by-step explanation:

if the bank charges 7.2% every year, and in this case it's 4 years, then that would mean it would be 28.8%(7.2 x 4), which means the answer would be 9000 x 28.8 which is 259200.

Hope it helped

3 0

3 years ago

Please help. Also please show your work. Y/5 - 2/5

Maurinko [17]

Answer:

0.2(y-2)

Step-by-step explanation:

factor out 1/5 from the expression

1/5×(y-2)

0.2(y-2)

8 0

2 years ago

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

Mr. Jones wants to buy 6 tickets to a play. Each ticket costs $10. Mr. Jones has $45 in his pocket.

lawyer [7]

Answer:

He needs $15 more

Step-by-step explanation:

since each ticket costs $10 dollars, and there is 6 tickets, we need to multiply 10 x 6 to get the total cost of all tickets

10 x 6 = $60

To find how much more money he needs to buy all tickets, we need to subtract $45 from the total cost of tickets, which is $60

$60 - $45 = $15

Hope you understand this explanation!

8 0

2 years ago