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3 years ago

Idk this question I need help

Mathematics

1 answer:

adell [148]3 years ago

8 0

To do it easily, you can divide the figure in 2: a rectangle and a triangle with 2 similar interior angles:
1: area of rectangle: 14 in × 6in
2: area of triangle:
to find the h: 19 -14= 5
to find the base: 2.5×2+6=11
area: (5×11)/2
as we havethe are of triangle and rectangle,we basically addthem together. in this we will findthe area of the entire figure.

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a box contains 20 blue marbes, 16 green marbles, and 14 red marbles. two marbles are selected at random. let 3 be the event that

Maksim231197 [3]

Answer:

Let E be the event that the first marble selected is green. Let F be the event that the second marble selected is green. A box contains 20 blue marbles, 16 green marbles and 14 red marbles P(F/E)=15/49 because if the first marble selected is green there are 49 in total and 15 are green. I think this is it.

Step-by-step explanation:

6 0

3 years ago

Make a frequency table to show the following test times (in minutes) for a reading test. 81, 63, 61, 58, 72, 70, 79, 68, 82, 64,

kodGreya [7K]

Sorry about my handwriting. this is what your graph should look like

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4 years ago

How many square centimeters are in an area of 5.60 in2?

Phantasy [73]

1 square inch = 2.54^2 = 6.4516 square cms

So 5.60 in^2 = 6.60 * 6.4516 = 36.13cm^2 to nearest hundredth

7 0

3 years ago

A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

TEA [102]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150 {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150 {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average balance = $1,200

s = sample standard deviation = $126

n = sample of account balances = 81

(b) So, test statistics = $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$ ~ $t_8_0$

= 3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0

4 years ago

14\4139 what is the answer

Masja [62]

Answer:

14 divided by 4139 is 0.0033824595312878

Step-by-step explanation:

7 0

3 years ago