ALEX HAD 45 TOY CARS.HE PUT 26 TOY CAR IN THE BOX. HOW MANY TOY CARS ARE NOT IN THE BOX ?

otez555 [7]

Answer:

given :-

for rectangular part: length=12in, breadth=8in

for triangular part:base=8in, height=3in

area of the given fig:

area of the given fig:area of 2 triangles +area of rectangle 

 $→2( \frac{1}{2} \times b \times h) + l \times b \\ →2( \frac{1}{2} \times 8 \times 3) + 12 \times 8 \\ →24 + 96 \\ → {120in}^{2}$ 

<h2>hope it helps you<3</h2>

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5 0

1 year ago

3.366 rounded to the nearest tenth of a ounce

olga nikolaevna [1]

Answer:

The answer is 3.4

Step-by-step explanation:

5 or more, round up

4 or less, round down

rounding 6

round up

so...

3.4

(-2, -7) and (-4, -9)

6 0

2 years ago

Read 2 more answers

Use technology to find the p-value for the hypothesis test described below. the claim is that for a smartphone carrier's data

Mila [183]

Given that the sample size is 12, thus the degree of freedom = 12 - 1 = 11.

Using technology, the p value of the t statistic, t = 2.028 is 0.9663.

8 0

3 years ago

Read 2 more answers

15% 40% as whole numbers

Alex17521 [72]

Answer: Not possible

Step-by-step explanation: A whole number is an integer or a number without fractions. Both 15% and 40% are less than 1 since they are less than 100%; therefore, it is impossible to express them as whole numbers. They can only be expressed as the mixed numbers of 0.15 and 0.4, or 15/100 and 40/100.

3 0

3 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

2 years ago