I don't understand this question so I need help

A recipe calls for the following spices: 18 teaspoon of turmeric, 18 teaspoon of ginger, and 14 teaspoon of cumin. What is the t

Sphinxa [80]

Answer:

50 total

Step-by-step explanation:

when it's asking for a total you add the numbers together, and whatever number you get is your answer :)

sorry if it's wronggg :( !!

7 0

3 years ago

I need help. this is due in less than 10 min and i have no clue how to solve. any tips appreciated

babymother [125]

Answers:

a) x = 9
b) arc JK = 68
c) arc MJ = 112
d) arc LMK = 248

================================

Explanation:

Arcs JK and KL form a semicircle, so they add to 180 degrees

(arcJK) + (arcKL) = 180

(5x+23) + (17x-41) = 180

22x-18 = 180

22x = 180+18

22x = 198

x = 198/22

x = 9

Then you'll use this x value to find arc JK and arc KL

arc JK = 5x+23 = 5*9+23 = 68

arc KL = 17x - 41 = 17*9-41 = 112

Since central angles MNJ and KNL are vertical angles, this means minor arcs MJ and KL are congruent arcs. So arc MJ is also 112 degrees

Arc LMK is basically nearly everything of the full circle, but we exclude out the portion from L to K (the shorter distance)

arc LMK = (full circle) - (measure of minor arc LK)

arc LMK = 360 - 112

arc LMK = 248

8 0

3 years ago

Can someone help “is 9 a factor of 23p - 47

quester [9]

Answer:

No.

Step-by-step explanation:

9 Does not go into 47 or 23

6 0

3 years ago

Read 2 more answers

V+6=92 solve for v

inysia [295]

Answer:

v = 86

Step-by-step explanation:

v + 6 = 92

-6 -6

v = 86

3 0

3 years ago

Read 2 more answers

To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t

nikklg [1K]

Answer:

$t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805$

$p_v =P(t_{(18)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

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

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

$\bar X_{1}=19.1$ represent the mean for the sample mean for 1

$\bar X_{2}=23.3$ represent the mean for the sample mean for 2

$s_{1}=4.818$ represent the sample standard deviation for the sample 1

$s_{2}=5.559$ represent the sample standard deviation for the sample 2

$n_{1}=10$ sample size selected 1

$n_{2}=10$ sample size selected 2

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1} \geq \mu_{2}$

Alternative hypothesis: $\mu_{1} < \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=10+10-2=18$

Since is a one sided test the p value would be:

$p_v =P(t_{(18)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

4 0

3 years ago