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3 years ago

Hhhhhhhellllllllp.....................................

Mathematics

1 answer:

Anit [1.1K]3 years ago

6 0

You have 28 meters of fence to make 3 sides.
28/3 = 9.3333

The pen will be 9.3333 by 9.3333 meters

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What single decimal multiplier would you use to increase by 9% followed by 3% decrease

MaRussiya [10]

Answer:

= 94

Step-by-step explanation:

To work out the multiplier, first add or subtract the percentage from 100, then convert to a decimal. Example: we want to add 20% to £110. To work out the multiplier, add 20 to 100, to get 120, and then change it to a decimal (divide by 100) to get 1.2. The answer is £110 x 1.2 = £132.

so to workout an increase of 9% followed by 3% we would do:

9% - 3% = 6%. convert = 0.06 = 100-6 = 94 converted into 0.94 x 100 = 94

7 0

3 years ago

Holly was given $5 000 by her parents for savings towards college. She invested part of the money in a government bond that paid

Elza [17]

Answer:

2500 x 1.06 + 2500 x 1.03

Step-by-step explanation:

2500 x 1.06 = 150

150 x .5 (half the simple interest) = 75

150 + 75 = 225

8 0

3 years ago

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

3 years ago

This is due today !! help pls

Anarel [89]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

PLS HELP I NEED TO FINISH THIS BEFORE TOMORROW

noname [10]

Answer: The first one, the last one,

Step-by-step explanation:

3 0

3 years ago