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tamaranim1 [39]

2 years ago

7

In 1 year, 120 students passed GCSE maths. The following year 98 passed. Find the percentage decrease in the pass rate. PLEASE P

LEASE PLEASE HELP QUICKLY!!!!!!20 PTS!!!!!

1 answer:

Margarita [4]2 years ago

3 0

Answer:

18% decrease.

Step-by-step explanation:

What you do is you do 120-98 to get 22. You then put that over 120 to get .18333 this equal to 18% decrease or if you need it specifically 18.3%

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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

2 years ago

Tyrone's bank account had a balance of -$65.87 last week. His paycheck of $349.59 was deposited this week. If there have been no

vladimir1956 [14]

What are the answers choices

4 0

2 years ago

What is the radius of the circle with this equation of (x-5)2+(y+7)2=8

AysviL [449]

Answer:

2√2

Step-by-step explanation:

compare this equation with the standard equation of a circle centered at (h, k) with radius r:

(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y + 7)^2=8

Clearly, r^2 = (radius)^2 = 8, and so the radius, r, is √8, or 2√2

3 0

2 years ago

zimovet [89]

.............................

7 0

1 year ago

A company that produces dog food claims that its bags contain 11 kg of food. From

andriy [413]

By testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Given mean of 10.6 kg, population standard deviation of 0.6 and sample size of 25.

We are required to find whether the company is right in saying that their bags contain 11 kg of food.

First we have to make hypothesis.

$H_{0}$ :μ≠11

$H_{1}$ :μ=11

We have to use t statistic because the sample size is less than 30.

t=(X-μ)/s/ $\sqrt{n}$

We will use s/ $\sqrt{n}$ =0.6 because we have already given population standard deviation of weights.

t=(11-10.6)/0.6

=0.4/0.6

=0.667

Degree of freedom=n-1

=25-1

=24

T critical at 0.05 with degree of freedom 24=2.0639

T critical at 0.05 with degree of freedom is greater than calculated t so we will accept the null hypothesis.It concludes that the bags donot contain 11 kg of food.

Hence by testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Learn more about t test at brainly.com/question/6589776

#SPJ1

3 0

1 year ago