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3 years ago

I need help with this word problem please help

Mathematics

2 answers:

alexandr402 [8]3 years ago

8 0

you will need 7 cups of sugar. 35 divided by 5 is 7.

antoniya [11.8K]3 years ago

7 0

Answer: You will need 7 cups of sugar.

Step-by-step explanation:

For every 5 eggs, you add one cup of sugar. 35 eggs divided by 5 eggs = 7 cups of sugar.

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(I) Factorice completely<br><br>x³- 6x² + 11x-6

adoni [48]

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Factor $x^3 - 6x^2 + 11x - 6$

$x^3 - 6x^2 + 11x - 6$

= $( x - 1) ( x - 2) ( x - 3)$

So you would be : $( x - 1) ( x - 2) ( x - 3)$

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If this helped you, could you maybe give brainliest..?

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8 0

4 years ago

A certain kind of investment requires that you pay a 10% penalty on any money you remove from the investment in the first 7 year

sergiy2304 [10]

Answer:

$22.80

Step-by-step explanation:

228 times 10% = 22.8 or $22.80

7 0

4 years ago

A Porsche uses 2.5 litres of fuel every 2 kilometres it travels. How much fuel does it use if it travels 50 kilometres

Sedbober [7]

2.5/2 = 1.25 litres per kilo
1.25 * 50 = 62.5 litres

Solution: 62.5 litres of fuel

8 0

3 years ago

Read 2 more answers

..........................

QveST [7]

Answer:

x = 0.954

Step-by-step explanation:

We apply pythagorean to find x

(4-x)² = 1.2² + 2.8²

(4-x)² = 9.28

✓(4-x)² = ✓(9.28)²

4 - x = 3.046

- x = 3.046 - 4

- x = -0.954

x = 0.954

(Please heart and rate if you find it helpful, it's a motivation for me to help more people)

7 0

3 years ago

A magazine reports that women trust recommendations from a particular social networking site more than recommendations from any

Misha Larkins [42]

Answer:

a) $p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

b) $p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size selected for male

c) $(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$

And the 95% confidence interval would be given (0.124;0.316).

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

$p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

Part b

$p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size elected for male

$z$ represent the critical value for the margin of error

Part c

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$

And the 95% confidence interval would be given (0.124;0.316).

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

8 0

3 years ago