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2 years ago

Please help with this 7th grade math I need to pass this test

Mathematics

2 answers:

kari74 [83]2 years ago

7 0

Area of triangle = (1/2) × Base × Height

Area of bigger triangle = (1/2) × 15 cm × 10 cm

Area of bigger triangle = (1/2) × 150 sq. cm

Area of bigger triangle = 75 sq. cm

Similarly,

Area of smaller triangle = (1/2) × 12 cm × 5 cm

Area of smaller triangle = (1/2) × 60 sq. cm

Area of smaller triangle = 30 sq. cm

Area of shaded region = Area of bigger triangle - Area of smaller triangle

Area of shaded region = 75 sq. cm - 30 sq. cm

Area of shaded region = 45 sq. cm

Answer :

Area of shaded region in sq. cm = 45

QveST [7]2 years ago

3 0

Answer is 45

Step by step exploration
A=10*15/2-5*12/2=75-30=45

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a fuel mixture consists of 3 parts oil and 11 parts gasoline. the oil costs $0.75 per liter, and the gasoline costs $1.60 per li

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Answer:

Total selling price= $663.82

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Step-by-step explanation:

Giving the following information:

Proportion of oil and gasoline:

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Gasoline= 11/14= 0.79

The oil costs $0.75 per liter.

The gasoline costs $1.60 per liter.

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3 years ago

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determinin

riadik2000 [5.3K]

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

$p=\frac{x}{n}$

$p=\frac{80}{100}$

$p=0.8$

Now we define hypothesis,

Null hypothesis $H_0$ : candidate A is significantly greater than 75%.

Alternative hypothesis $H_1$ : candidate A is not significantly greater than 75%.

Level of significance $\alpha=0.05$

Applying test statistic Z -proportion,

$Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Where, $\widehat{p}=80\%=0.80$ and $p=75%=0.75$

Substitute the values,

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}$

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}$

$Z=\frac{0.05}{0.0433}$

$Z=1.1547$

The p-value is

$P(Z>1.1547)=1-P(Z$

$P(Z>1.1547)=1-0.8789$

$P(Z>1.1547)=0.1241$

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

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