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natulia [17]
3 years ago
9

Six cakes cost £1.80 how much do ten cakes cost

Mathematics
2 answers:
likoan [24]3 years ago
4 0
Here, 6 cakes cost = £1.80
So, 1 cake costs = 1.80/6

Now, for 10 cakes, it would be: 1.80/6 * 10 = 18/6 = 3

In short, Your Final Answer would be £3

Hope this helps!

julia-pushkina [17]3 years ago
3 0
£1.80 / 6 = 30p
this is the cost of 1 cake
30p x 10 = £3
the cost of 10 cakes is £3

hope this helps you
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pochemuha

The volume of a shape is the amount of space in it.

<em>The volume outside the cylinder, but inside the cube is 214.6 cubic inches</em>

The given parameter is:

l = 10 ---- length of the cube

So, the volume of the cube is:

V_1 = l^3

V_1 = 10^3

V_1 = 1000

The volume of the cylinder is:

V_2 = \pi r^2h

The diameter (d) of the cylinder equals the length of the cube.

This means that:

d = l

d = 10

Divide by 2 to calculate the radius of the cylinder

r = \frac{10}{2}

r = 5

Also, the height of the cube is::

h =l

h =10

So, we have:

V_2 = \pi r^2h

V_2 = \pi \times 5^2 \times 10

V_2 = \pi \times 25 \times 10

V_2 = \pi \times 250

V_2 = 785.4

So, the volume (V) outside the cylinder is:

V = V_1 - V_2

V = 1000 - 785.4

V = 214.6

Hence, the volume outside the cylinder is 214.6 cubic inches

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brainly.com/question/15861918

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2 years ago
What is x ???? help me please
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Answer:

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Step-by-step explanation:

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Answer:

a. The conditions are met to use a large-sample confidence interval.

b. [0.048; 0.171]

c. Decision: Not reject the null hypothesis.

d. Check explanation.

Step-by-step explanation:

Hello!

You have the following study variable:

X: red-eyed fruit fly that shows to be heterozygous due to producing mixed progeny after being crossed with s white-eyed fruit fly.

n= 100

x= 11

sample proportion 'p= 0.11

a)

Binomial criteria:

1. The number of observation of the trial is fixed (In this case n = 100)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial

3. The probability of success in the same from one trial to another (In this case our "success" is that the fruit fly show heterozygous trough its progeny)

So X≈ Bi (n;ρ)

Considering that the sample is big enough (n≥30), you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal. So you can use the large-sample confidence interval.

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'p ± Z_{1-\alpha /2}*\sqrt{\frac{'p(1-'p)}{n} }

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c.

H₀: ρ = 0.10

H₁: ρ ≠ 0.10

α: 0.05

Z= \frac{'p - p}{\sqrt{\frac{p(1-p)}{n} } }

Z= \frac{0.11 - 0.10}{\sqrt{\frac{0.10*0.90}{100} } }

Z= 0.33

The two tailed p-value for this test is 0.7414. The p-value is greater than the level of significance so the decision is to not reject the null hypothesis.

d.

To be able to compare a Confidence Interval several conditions shoud be met.

1) The interval and the hypothesis test should be made for the sale population parameter.

2) The hypothesis has to be two-tailed

3) Both levels (confidence and significance) should be complementary.

4) Both have to be made with the information of the same sample. (Remember that the variable is random and the values will change from sample to sample therefore it makes no sense to compare an interval and hypothesis of different samples. You could reach a wrong conclusion)

If the conditions are met, you check the calculated interval to see if it contains the value of the parameter under the null hypothesis. If the interval contains the value (In this case 0.10) then the hypothesis is to be supported. If the interval doesn't include the value, then the hypothesis is to be rejected.

I hope this helps!

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