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3 years ago

Judy can decorate 3 cakes in 5 hours. Which graph represents this same rate?

Mathematics

2 answers:

Gnoma [55]3 years ago

5 0

Answer:

A. the one with pink because it goes through the 3 and 5 mark.

Step-by-step explanation:

mojhsa [17]3 years ago

3 0

Answer:

The first one with the pink background

Step-by-step explanation:

It's the only one that matches perfectly

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PLEASE HELP ASAP! WILL GIVE BRAINLEIST! Find the area of the shaded region. Use pi = 3.14

Semmy [17]

Answer:

Step-by-step explanation:

Square

The side length of the square = 8 cm

The area of the square = s^2

The area of the square = 8^2 = 64

Circle

r = d/2

r = 8/2

r = 4

Area of the circle = pi r^2

Area of the circle = 3.14 * 16

Area of the circle = 50.24

Shaded Area

Area of the Shaded area = area Square - Area Circle

= 64 - 40.24

= 23.76

4 0

3 years ago

Twenty-two is what percent of 55 A) 22% B) 40% C) 50% D) 250%

Mashutka [201]

Answer:

twenty two percent of 55 would be the answer B.40 I think.

4 0

3 years ago

What is the expanded form of 603478

WINSTONCH [101]

600000+3000+400+70+4

4 0

3 years ago

Read 2 more answers

AT&T:$0.08 per minute,$5 monthly fee VOIP:(cable phone):free calls,$35 monthly fee Cell:$40 for first 400 minutes,$0.25 per

Luden [163]

Make 2 graphs and with the second graph graph the cell fee and with the first Do the minutes

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3 years ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

3 years ago