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2 years ago

What is the area of the trapezoid shown below?

Mathematics

1 answer:

andrew-mc [135]2 years ago

5 0

Ok so the triangle part is 10 but I don’t know what the rectangle part is but the formula for a rectangle is A=base x high

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Allison has five times as many baseball cards as football cards and all she has 120 baseball and football cards how many basebal

Shtirlitz [24]

The answer is 600 because you have to multiply 120 by 5 and the sum is 600

7 0

3 years ago

1. What is the common

uysha [10]

Answer:

Step-by-step explanation:

It is add three add three add three add three

5 0

2 years ago

Pls ASAP help me with number 14

attashe74 [19]

Answer:

D. 19

Step-by-step explanation:

because they are parallel bases you can compare

63/57 = 21/x

x = 19

4 0

2 years ago

Read 2 more answers

If 32^2c=8^c what is the value of c? 1 2 3 5

Elina [12.6K]

Answer:

as written, c ≈ 0.000979 or c = 4
alternate interpretation: c = 0

Step-by-step explanation:

<em>As written</em>, you have an equation that cannot be solved algebraically.

(32^2)c = 8^c

1024c = 8^c

1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero

A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.

___

If you mean ...

32^(2c) = 8^c

(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2

2^(10c) = 2^(3c) . . . . . . . simplify

10c = 3c . . . . . . . . . . . . . .log base 2

7c = 0 . . . . . . . . . . . . . . . subtract 3c

c = 0 . . . . . . . . . . . . . . . . divide by 7

7 0

2 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago