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2 years ago

14

On Friday night, Alex ate a large pizza for dinner. He had 3/5 of the pizza left over when he was done. On Saturday, he ate 1/2

of the leftover pizza from Friday. How much pizza did Alex eat on Saturday?

1 answer:

ad-work [718]2 years ago

5 0

Answer:

3/10

Step-by-step explanation:

3/5 left over and he eats half of it. So 3/5 ×1)2 which is 3/10.

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If the common difference of an ap is 3/2 and its 20 th term 35×1/2 find first term and 15 th term

Lady bird [3.3K]

Answer:

Step-by-step explanation:

d = 3/2

a₂₀ = a₁+19d

35/2 = a₁ + 19×3/2

a₁ = 35/2 - 19×3/2 = -11

a₁₅ = a₁+14d = -11 + 14×3/2 = 10

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3 years ago

Jason invests $1000 at a bank that offers 3% compounded annually. Write an equation to model the growth of the investment

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3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

diego is collecting pennies and estimated that he has 120 pennies. he counted them and has 80 pennies. what is diego's percentag

Bezzdna [24]

Answer:

50%

Step-by-step explanation:

The percent error is equal to the absolute value, difference of the measured value or your experimental value minus the actual value divided by the actual value times 100%;

Here we have;

Measured Valued = 120 pennies

Actual Value = 80 pennies

% error = $\frac{|Measured - Actual|}{Actual} * 100$

% error = $\frac{|120 - 80|}{80} * 100$

% error = $\frac{|40|}{80} * 100$

% error = $\frac{40}{80} * 100$

% error = 50%

So, 50% is the percent error for Diego

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3 years ago

What is the slope of the line

kolbaska11 [484]

Answer:

B.

Step-by-step explanation: Because when you count them down to 3/5 It's more in the negative.

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3 years ago

Read 2 more answers