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3 years ago

Answer all questions please

Mathematics

1 answer:

Marat540 [252]3 years ago

4 0

Answer:

a. base angle

b. leg

c. vertex

d. leg

e. base

f. base

5. You have 3 pairs of corresponding congruent sides, so the property that proves the triangles are congruent would be SSS.

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753% in a decimal form<br> :)

sasho [114]

ANSWER: 7.53

Step-by-step explanation:

To convert a percent to decimal you need to divide the percent by 1,

753% ÷ 1 = 7.53

7 0

3 years ago

Express 0.035 metre into km

azamat

Answer:

$0.000035km$

Step-by-step explanation:

$1km =1000m \\ x = 0.035m$

$use \: \: cross \: \: multipication$

$0.035 \times 1 = 1000x \\ 0.035 = 1000x \\ \frac{0.035}{1000} = \frac{1000x}{1000} \\ x = 0.000035km$

6 0

3 years ago

Read 2 more answers

Find the value of the combination. 10C0

Andrej [43]

Answer:The formula for a combination of choosing r unique ways from n possibilities is:

nCr = n!

r!(n - r)!

where n is the number of items and r is the unique arrangements.

Plugging in our numbers of n = 10 and r = 0, we get

10C0 = 10!

0!(10 - 0)!

10C0 = 3,628,800

3,628,800

8 0

1 year ago

Read 2 more answers

Which of the following is not a rate?

IrinaK [193]

Answer:

Folding 6 loads

Step-by-step explanation:

5 0

3 years ago

An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing

wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1 $\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

$S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

$S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

$S=0.9905$

Generally the equation for test statistics is mathematically given by

$t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

$t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$t=-4.00091$

Therefore From table

$P value=P(t_{15+15-2}=4.00091)$

$P value=0.00041918$

$P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

3 0

2 years ago