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2 years ago

Help me out please !!!

Mathematics

1 answer:

Nadya [2.5K]2 years ago

8 0

Answer:

rate of change = 1.75 over 1 or just 1.75

Step-by-step explanation:

this represents the cost of each cantaloupe. This basically means that each cantaloupe is $1.75.

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12a+6b equivalent expression

patriot [66]

Answer:

$6(2a+b)$

Step-by-step explanation:

The given expression is

$12a+6b$

To find the equivalent expression of the given one, we just have to factor.

In this case, the only thing we can do to rewrite is to extract the common greatest factor, which is 6

$12a+6b\\6(2a+b)$

Therefore, the equivalent expression is

$6(2a+b)$

5 0

3 years ago

Please tell me answer of 6 7 and 8

zloy xaker [14]

Answer:

6.angle m=180-60=120

7.angle y=180-(60+65)=55

8 0

2 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - α) % confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error in this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

2 years ago

Find the GCF of the terms of the polynomial.

-BARSIC- [3]

<h3>Answer: 2z^3</h3>

====================================================

Explanation:

The GCF of the coefficients {6, -42, 14} is 2 as it is the largest factor found in each of the three values.

----------

For the variable portions, we can

write z^5 as z^3*z^2
write z^4 as z^3*z^1
write z^3 as z^3*1

Each time we see z^3 show up, so this is the largest common factor among the three variable terms.

----------

The two results we got were 2 and z^3

Putting the two results together, we end up with the overall GCF of 2z^3

4 0

3 years ago

Find the length of __ JK Round answer to nearest tenth.

PIT_PIT [208]

Answer:

Step-by-step explanation:

We would apply the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and A is the angle corresponding to a. Likening the expression to the given triangle, it becomes

JK² = JL² + KL² - 2(JL × KL)Cos10

JK² = 61² + 53² - 2(61 × 53)Cos10

JK² = 3721 + 2809 - 6466Cos10

JK² = 6530 - 6367.767

JK² = 162.233

Taking square root of both sides of the equation, it becomes

JK = √162.233

JK = 12.73 to the nearest tenth

6 0

2 years ago