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Andrew [12]
3 years ago
11

Diep buys a loaf of bread 65 centimeters long. For lunch every afternoon, he cuts 15 centimeters of bread for his sandwich. Diep

wants to determine the length of the loaf of bread, l, after d days. What is the equation of the scenario? Is the graph of the equation continuous or discrete? a) l = 65 – 15d; discrete b) l = 65 – 15d; continuous c) 65 = l – 15d; discrete d) 65 = l – 15d; continuous
Mathematics
2 answers:
Vladimir [108]3 years ago
7 0

Answer:

a

Step-by-step explanation:

After d days, the loaf of bread is 65 - 15d long.

l = 65 - 15d.

The graph is discrete, as he cannot cut the bread for 1.5 days, or cut 5cm of the bread off.

ki77a [65]3 years ago
6 0

Answer:

l = 65 – 15d; discrete

Step-by-step explanation:

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Rebecca bought candy boxes in the shape of a square pyramid to use for party favors. The side length of each box is 3.6 inches,
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Answer:

yes she will

Step-by-step explanation:

im not very sure if it is correct, and if its wrong then im sorry

4 0
3 years ago
Read 2 more answers
2+7/9x=5/6 Rewrite without fractions or decimals
quester [9]

Answer:

2 + (7 ÷ 9)x = 5 ÷ 6

Step-by-step explanation:

Given

2+\frac{7}{9}x=\frac{5}{6}

You can eliminate fractions by turning them into division problems!

2 + (7 ÷ 9)x = 5 ÷ 6

5 0
2 years ago
5 out of 20 students said ice cream was their favorite snack if they were a hundred six graders how many would chose ice cream P
Rufina [12.5K]

Answer:

<u>25 students</u>

Step-by-step explanation:

20 times 5 equals 100

5 times 5 equals 25

To get to 100 students you would times it by 5 and because we times'd the 20 times 5 we need to do the same to the 5 which would result in 25 students. Hope this helped! xx

4 0
3 years ago
A county environmental agency suspects that the fish in a particular polluted lake have elevated mercury levels. To confirm that
suter [353]

Answer:

a. The 95% confidence interval for the difference between means is (0.071, 0.389).

b. There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

In the case of the confidence interval, we reach this conclusion because the lower bound is greater than 0. This indicates that, with more than 95% confidence, we can tell that the difference in mercury levels is positive.

In the case of the hypothesis test, we conclude that because the P-value indicates there is a little chance we get that samples if there is no significant difference between the mercury levels. This indicates that the values of mercury in the polluted lake are significantly higher than the unpolluted lake.

Step-by-step explanation:

The table with the data is:

Sample 1 Sample 2

0.580    0.382

0.711      0.276

0.571     0.570

0.666    0.366

0.598

The mean and standard deviation for sample 1 are:

M=\dfrac{1}{5}\sum_{i=1}^{5}(0.58+0.711+0.571+0.666+0.598)\\\\\\ M=\dfrac{3.126}{5}=0.63

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.58-(0.63))^2+...+(0.598-(0.63))^2]}\\\\\\            s=\sqrt{\dfrac{1}{4}\cdot [(0.002)+(0.007)+(0.003)+(0.002)+(0.001)]}\\\\\\            s=\sqrt{\dfrac{0.015}{4}}=\sqrt{0.0037}\\\\\\s=0.061

The mean and standard deviation for sample 2 are:

M=\dfrac{1}{4}\sum_{i=1}^{4}(0.382+0.276+0.57+0.366)\\\\\\ M=\dfrac{1.594}{4}=0.4

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(0.382-(0.4))^2+(0.276-(0.4))^2+(0.57-(0.4))^2+(0.366-(0.4))^2]}\\\\\\            s=\sqrt{\dfrac{1}{3}\cdot [(0)+(0.015)+(0.029)+(0.001)]}\\\\\\            s=\sqrt{\dfrac{0.046}{3}}=\sqrt{0.015}\\\\\\s=0.123

<u>Confidence interval</u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

M_d=M_1-M_2=0.63-0.4=0.23

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07

The critical t-value for a 95% confidence interval is t=2.365.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=2.365 \cdot 0.07=0.159

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 0.23-0.159=0.071\\\\UL=M_d+t \cdot s_{M_d} = 0.23+0.159=0.389

The 95% confidence interval for the difference between means is (0.071, 0.389).

<u>Hypothesis test</u>

This is a hypothesis test for the difference between populations means.

The claim is that the fish in this particular polluted lake have signficantly elevated mercury levels.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

The significance level is 0.05.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

M_d=M_1-M_2=0.63-0.4=0.23

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.23-0}{0.07}=\dfrac{0.23}{0.07}=3.42

The degrees of freedom for this test are:

df=n_1+n_2-1=5+4-2=7

This test is a right-tailed test, with 7 degrees of freedom and t=3.42, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>3.42)=0.006

As the P-value (0.006) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

<u> </u>

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

In the case of the confidence interval, we reach this conclusion because the lower bound is greater than 0. This indicates that, with more than 95% confidence, we can tell that the difference in mercury levels is positive.

In the case of the hypothesis test, we conclude that because the P-value indicates there is a little chance we get that samples if there is no significant difference between the mercury levels. This indicates that the values of mercury in the polluted lake are significantly higher than the unpolluted lake.

7 0
3 years ago
Sketch the graph that has an area of under the standard normal curve and to the left of the​ z-score. Then find the associated​
Anestetic [448]

The normal distribution is also known as the Gaussian distribution. The correct option is B, while the approx z-score is 0.31.

<h3>What is Normal Distribution?</h3>

The normal distribution, also known as the Gaussian distribution, is a symmetric probability distribution about the mean, indicating that data near the mean occur more frequently than data distant from the mean. The normal distribution will show as a bell curve on a graph.

Since the normal distribution is symmetrical about the mean. Therefore, the correct option is B, while the approx z-score is 0.31.

Learn more about Normal Distribution:

brainly.com/question/15103234

#SPJ1

4 0
2 years ago
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