All categories

3 years ago

8k-1=15 ? Help I have a question were we have to work this out what do I do?

2 answers:

7 0

You have to add 1 to the -1 and you will get 0.What you have to do next is also add 1 to 15 because you want a variable on one side and a number on the other.So now you have 8k=16. So you divide 8 to 8k and you get nothing divide 8 and 16 and your answer for k =2 hope you get it right.trust mw it is very easy when you get use to it

adell [148]3 years ago

5 0

$8k-1=15\ \ \ \ /+1\\\\8k-1+1=15+1\\\\8k=16\ \ \ \ /:8\\\\8k:8=16:8\\\\k=2$

You might be interested in

%) 1.3.27

jeka94

Answer:

Width of field = 50 yards

Length of Yards = 106 yards

Step-by-step explanation:

Let the width of field be x yards

given that "The length of a new rectangular playing field is 6 yards longer than double the width"

Length of Yards= 2*width of field + 6 = 2x+6 yards

perimeter of the rectangular playing field = 312 yards

We know that perimeter of rectangle is given by 2(length + width ).

Since field is rectangular then

2(length + width ) = 312

2(x+2x+6) = 312

=> 3x+6 = 312/2 = 156

=> 3x = 156 - 6 = 150

=> x = 150/3 = 50

Thus,

Width of field is x = 50 yards

Length of Yards= 2x+6 yards = 2*50+6 yards = 106 yards

5 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

Easy anwer!!!!!!Will mark brainliest

elena55 [62]

Answer:

95degrees

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Factor: 125x^6-75x^4-20x^2+12

Naddika [18.5K]

Answer:

( 5 2 − 3 ) ( 5 2 − 2 ) ( 5 2 + 2 )

Step-by-step explanation:

8 0

3 years ago

20-10 write four equivalent expressions

igor_vitrenko [27]

Here's two. I can't think of any more.

20/10
20-(-10)

5 0

3 years ago