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3 years ago

How many FIVES are equal in value to £7?

1 answer:

5 0

£1 equals to 1.43 us dollars.

7 × 1.43 = 10.01

10.01 ÷ 5 = 2.002

2 fives are equal to £7.

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(0.2c^3)^2–0.01c^4(4c^2–100)

Debora [2.8K]

The answer should be C^4

3 0

3 years ago

Read 2 more answers

The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .

sammy [17]

Answer:

$32,24,18,\frac{27}{2} ,\frac{81}{8}$

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is $32,x,y,z,\frac{81}{8}$

Recall that term of a geometric sequence are generally in the form:

$a,ar,ar^2,ar^3,ar^4$

This implies that:

a=32 and $ar^4=\frac{81}{8}$

Substitute a=32 and solve for r.

$32r^3=\frac{81}{8}$

$r^4=\frac{81}{256}$

Take the fourth root to get:

$r=\sqrt[4]{\frac{81}{256} }$

$r=\frac{3}{4}$

Therefore $x=32*\frac{3}{4} =24$

$y=24*\frac{3}{4} =18$

$z=18*\frac{3}{4} =\frac{27}{2}$

8 0

3 years ago

16=5y/3 HELLLPPP PLSSSS

Lunna [17]

Answer:

y = 9.6

Step-by-step explanation:

mark brainliest

4 0

3 years ago

PLEASE HELP WITH MATH! 100 PTS

timurjin [86]

Answer:

1.) step No.3 x(2x-5) - (2x-5)

2.) 16 +28i-28i - 49i²

but we know i² = -1

the expression becomes

16 + 49

= 66

6 0

3 years ago

Read 2 more answers

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 98% conf

adelina 88 [10]

Answer:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1.1$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

3 0

3 years ago