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3 years ago

Pleaseee helpp im giving a lot of points if corrected

Mathematics

1 answer:

Dafna11 [192]3 years ago

3 0

Answer:

x = 38°

Step-by-step explanation:

Complementary angles = 90°

90° = x + x+14

90° = 2x + 14

104 = 2x

x = 52°

the next angle should be 52 - 14 = 38°

So x = 38°

If my answer is incorrect, pls correct me!

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-Chetan K

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AKAJAXOSPAPXODAOAPA HELP

Leya [2.2K]

Answer:

He used 6 bananas at .50 which cost $3.00

12 oranges at .75 which cost $9.00 and

3 papayas at $1.25 which cost $3.75

Step-by-step explanation:

We can set up a system of equations using the information given.

b for banana, r for orange, p for papaya

b + r + p = 21. Also given r = 2b So substitute to simplify for calculating two unknowns:

b + 2b + p = 21 The number of items.

50b + 75(2b) +125p = 1575 The cost.

I multiplied everything by 100 to eliminate the decimals.

To solve by substituion, rewrite the first equation to get a value for p in terms of b

p = 21 - 3b

Substitute this value in the second equation and solve for b.

50b + 150b + 125(21–3b) = 1575

200b +2625 –375b = 1575 Subtract 2625 from both sides and combine the like terms:

–175b = 1575 –2625

–175b = –1050. Divide both sides by –175

b = 6

Substitute in the original equation to find the amount of each fruit:

b + r + p = 21

b = 6 bananas

r = 2b. 2(6) = 12 oranges. I used r so as not to mix up o with 0.

6 + 12 + p = 21

p = 21 –18

p = 3 papayas

7 0

3 years ago

572 to the nearest hundred

levacccp [35]

72 is 50 or above so you round it up.
572 to the nearest hundred is 600

3 0

3 years ago

Read 2 more answers

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:

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

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