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3 years ago

8

Lucy and her mother made tacos they put 2 tacos on each of 7 plates select the number sentences that show all the tacos Lucy and

her mother made mark all that apply 2+7=9. 7+7=14. 8+6=14. 2 times 7=14

2 answers:

vesna_86 [32]3 years ago

7 0

2 times 7 = 14 is the right answer

Mashutka [201]3 years ago

6 0

3(7)=14 is the correct answer

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a random sample of 510 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2

34kurt

Answer:

CI = 21 ± 0.365

Step-by-step explanation:

The confidence interval is:

CI = x ± SE * CV

where x is the sample mean, SE is the standard error, and CV is the critical value (either t score or z score).

Here, x = 21.

The standard error for a sample mean is:

SE = σ / √n

SE = 3.2 / √510

SE = 0.142

The critical value is looked up in a table or found with a calculator. But first, we must find the alpha level and the critical probability.

α = 1 - 0.99 = 0.01

p* = 1 - (α/2) = 1 - (0.01/2) = 0.995

Using a calculator or a z-score table:

P(x<z) = 0.995

z = 2.576

Therefore:

CI = 21 ± 0.142 × 2.576

CI = 21 ± 0.365

Round as needed.

8 0

3 years ago

Two different crews earn different amounts for different hours they work.

Liula [17]

Answer:

I need to see the graph in order to answer the question

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

In the lab, Leila has two solutions that contain alcohol and is mixing them with each other. Solution A is alcohol and Solution

torisob [31]

Answer:

300ml

Step-by-step explanation:

As values are missing in your question, the complete question is:

In the lab, Leila has two solutions that contain alcohol and is mixing them with each other. Solution A is 6% alcohol and Solution B is 20% alcohol. She uses 400 milliliters of Solution A. How many milliliters of Solution B does she use, if the resulting mixture is a 12% alcohol solution?

Considering x=ml of 20% solution B

Therefore, 400+x=ml of resulting 12% solution

Solution A alcohol+ solution B alcohol= Alcohol solution

6% (400)+ 20%x = 12%(400+x) ->(converting percentage into decimal)

.06*400+.2x=.12(400+x)

24+.2x=48+.12x

.08x=24

x=24/.08=300 ml

she used 300 milliliters of 20%Solution B in resulting mixture.

4 0

2 years ago

I will give Brainliest!!!

mixer [17]

Answer: C

Step-by-step explanation:

x = 2 $\sqrt{y+2}$ - 6

x + 6 = 2 $\sqrt{y+2}$

((x+6)/2 = $\sqrt{y+2}$ )^2

(x+6/2)^2 = y+2

y = (x+6/2)^2 -2

y= 1/4x^2 +3x +7

3 0

2 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$

Result is not significant.

3.

$(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378$

Result is not significant.

4.

$(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349$

Result is not significant.

5.

$(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773$

Result is not significant.

6.

$(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0$

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair die.

8 0

2 years ago