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3 years ago

Can someone help please?

Mathematics

1 answer:

ratelena [41]3 years ago

4 0

Answer: It is TRUE. A point in a linear equation can be the solution to said equation.

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Martin has 5/6 of a pizza. Liam has 4 times as much pizza as Martin.

Novosadov [1.4K]

Answer:

Liam has 3 whole pizza along with 1/3 of a pizza.

In decimal, Liam has 3.33 of Pizza.

Step-by-step explanation:

Given

Quantity of pizza with Martin = 5/6 of a pizza

Pizza with Liam = 4*Quantity of pizza with Martin

Pizza with Liam = 4*5/6 of a pizza

Pizza with Liam = 10/3 of a pizza = 3 1/3 of pizza.

Thus, Liam has 3 whole pizza along with 1/3 of a pizza.

In decimal Liam has 3.33 of Pizza.

7 0

3 years ago

Eva has $10 for lunch. Her meal cost $8.60. She wants to leave a 20% tip. Does she have enough money? Explain your reasoning.

snow_tiger [21]

No she doesn't have enough money. All you do is multiply 8.60 by .2 which come out to 1.72 and 8.60 plus 1.73 which equals 10.33 so she needs 33 more cents tone able to tip 20%.

6 0

3 years ago

What is the value of ( 4^2 − 6 ) ÷ 2 × 3^2 ?

Amanda [17]

Answer: 45

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Leon got a reduction of $40 on a Geometry set

SSSSS [86.1K]

I pretty sure this one is b

6 0

2 years ago

Read 2 more answers

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of <em>z</em> for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

3 years ago