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rosijanka [135]

3 years ago

7

I need help with #29 I really don't understand it (picture included)

1 answer:

Katena32 [7]3 years ago

6 0

4x+4(3)+4(2)=y
4x+20=y
X equals the amount it cost for each ticket which is unknown. Y equals the total cost. Because the amount of the food is provided, it can be calculated unlike the tickets which is just left at 4x. The work for the snacks are upwards. Its mostly substitution.

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A local hardware store sold 12 percent of its hammers on Saturday . If the hardware store sold 9 hammers on Saturday how many ha

Dmitrij [34]

$\frac{12}{100}$ = $\frac{9}{x}$ Cross multiply

12x=900 Divide both sides by 12.

x=75 hammers to begin with.

7 0

3 years ago

Does anyone know how to do this

Scilla [17]

Answer:

$7 {x}^{2} + 28x - 35 = 0$

$7 {x}^{2} + 35x - 7x - 35 = 0$

$7x(x + 5) - 7(x + 5) = 0$

$(x + 5)(7x - 7) = 0$

$(x + 5)(x - 1) = 0$

3 0

3 years ago

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A science fair poster is a rectangle 3 feet long and 2 feet wide. What is the area of the poster in square inches? Be sure to in

vredina [299]

Area = length · width
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hope this helps!

5 0

3 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

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

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

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

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

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

3 years ago

Anybody plz help me will give brainliest!!!

kodGreya [7K]

Answer:

24

Step-by-step explanation:

6 0

3 years ago

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