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svet-max [94.6K]

3 years ago

11

Joy ate 1/4 of a pizza. If she divides the rest of the pizza into pieces equal to 1/8 pizza for her family, how many pieces will

her family get? how did you get the 6? Thank you

1 answer:

Ann [662]3 years ago

4 0

3/4 ×2/2 to get the denominator to 8

You might be interested in

Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.50.

Mariulka [41]

Answer:

95% Confidence interval for taxi fare: ($20.5,$24.2)

Step-by-step explanation:

We are given the following data set: for fares:

$22.10, $23.25, $21.35, $24.50, $21.90, $20.75, and $22.65

Formula:

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{156.5}{7} = 22.35$

95% Confidence interval:

$\bar{x} \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$22.35 \pm 1.96(\frac{2.5}{\sqrt{7}} ) = 22.35 \pm 1.85 = (20.5,24.2)$

6 0

3 years ago

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0

3 years ago

Thank you for you help

Harrizon [31]

the first simple interest will be $1.73 more than the second one.

<h3>Which is the difference between the two interests?</h3>

The loan is of $575, and there are two options:

7 years at 4.5% or 8 years at 3.9%

Remember that the simple interest formula is:

SI = (P*R*T)/100

Where:

P = principal value.

R = rate (this is the percentage per year)

T = time, in years.

Then for the first option. the interest will be:

SI = ($575*4.5*7)/100 = $181.13

For the second option we will have the interest:

SI' =( $575*3.9*8)/100 = $179.4

The difference is:

$181.13 - $179.4 = $1.73

This means that the first simple interest will be $1.73 more than the second one.

If you want to learn more about simple interests:

brainly.com/question/20690803

#SPJ1

7 0

1 year ago

Solve the inequality 2x - 3

Crank

That's not an inequality.

6 0

3 years ago

I used 2/3 for tickits and 1/6 on popcorn how much money is left

posledela

Answer:

You would have 1/6 of your money remaining.

Step-by-step explanation:

Let's write an equation for this.

1 - (2/3 + 1/6)

First, let's multiply the numerator and denominator of 2/3. You get 4/6!

1 - (2/3 + 1/6)

1 - (4/6 + 1/6)

Next, let's add 4/6 and 1/6. You get 5/6!

1 - (4/6 + 1/6)

1 - 5/6

Finally, let's turn 1 into 6/6 and do 6/6 - 5/6!

1 - 5/6

6/6 - 5/6

1/6

You would have 1/6 of your money remaining.

3 0

3 years ago