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Phoenix [80]
3 years ago
10

What is the two summaries for this question

Mathematics
1 answer:
umka2103 [35]3 years ago
5 0
That's not a question
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Which of these is an equation? 8t+4=76 8t+4 8z 8+4
vovangra [49]

Answer:

8t+4=76

Step-by-step explanation:

An equation needs an equals sign otherwise it is an expression

4 0
3 years ago
Amanda Sabino's taxable income is $20,900 . Use this table below to find out how much she'll pay in state tax
Darina [25.2K]

Answer:

is that a test huh

is that a test

Step-by-step explanation:

8 0
3 years ago
The measure of EC bisects
stepladder [879]

Answer:

The value of x = 20

Step-by-step explanation:

Given:

∠BOC = 3x - 15

∠COD = 2x + 5

Find:

The value of x

Computation:

⇔ ∠BOC + ∠COD = 90

⇔ 3x - 15 + 2x + 5 = 90

⇔ 5x - 10 = 900

⇔ 5x = 90 + 10

⇔ 5x = 100

⇔ x = 100 / 5

⇔ x = 20

The value of x = 20

5 0
3 years ago
Sophia wants to enlarge a 5-inch by 7-inch rectangular photo by multiplying the dimensions by 3. Find the area of the enlarged p
Mama L [17]

Answer:

315 sq. in

Step-by-step explanation:

15 * 21 = 315

8 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 - <em>α</em>)% 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:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 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
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