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2 years ago

Help me please it's easy ig

Mathematics

1 answer:

Gelneren [198K]2 years ago

8 0

Answer: 9 R53

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Find the solution to the system of equations.

N76 [4]

Answer:

x = -2, y = 5.

Step-by-step explanation:

y=−4x−3

y=−2x+1

Because y is common to both left sides:

-4x - 3 = -2x + 1

-4x + 2x = 1 + 3

-2x = 4

x = 4/-2

x = -2.

So y = -4(-2) - 3

= 8 - 3

= 5 .

5 0

2 years ago

Mount Kilimanjaro is approximately 19,000 feet in height. as you go up in altitude through the atmosphere near earth surface the

horsena [70]

Answer:

10℉

Step-by-step explanation:

9,000 feet - 4000 feet (because the camp is 4000 feet above sea level) = 5000 feet

for every 1000 ft you decrease by 3.6℉ therefore 5(3.6) = 18 degrees

when you get to the top it is -8℉ to find out the temp at base camp you just add 18 ℉ to the final tempreture

-8 +18 = 10℉

5 0

3 years ago

Martha breaded 1/2 of fish and ate 1/3 how much did Martha eat

tigry1 [53]

So she breaded 1/2 of the fish and then ate 1/3 of that 1/2. So we need to find 1/3 of 1/2.

When you see "of" it means "×"

1/3 of 1/2 =
1/3 × 1/2 =
1/6

She ate 1/6 of the fish. She only ate 1/6 because fish is gross.

6 0

3 years ago

In a certain month a family spent $750 on food .In third month the family spent $729 on food .What was percentage decrease in th

seraphim [82]

Answer: 2.8%

Step-by-step explanation:

Give, In a certain month a family spent $750 on food

In third month the family spent $729 on food .

i.e. Expenditure on food in a certain month = $750

Expenditure on food in third month = $729

Decrease in monthly spending on food = (Expenditure in a certain month )- (Expenditure in third month )

= $750 - $729

= $21

Now , the percentage decrease in the month spent on food will be :

$\dfrac{\text{Decrease in monthly spending on food }}{\text{Expenditure on food in a certain month}}\times100\\\\=\dfrac{21}{750}\times100\\\\=2.8\%$

Hence, the percentage decrease in the month spent on food is 2.8%.

3 0

3 years ago

Consider testing Upper H 0 : mu equals 20H0: μ=20 against Upper H Subscript a Baseline : mu less than 20Ha: μ<20 where muμ is

kolezko [41]

Answer:

Test: $\mu$ < 20 with normal distⁿ

Hypothesis test:

H0: $\mu \ge$ 20 (Null Hypothesis, H₀ : $\mu$ = 20 is also correct)

H1: $\mu$ < 20 (Alternative Hypothesis, also called H₁)

This is lower tailed test.

Since sample is large, sample standard deviation can be taken as an approximation of population standard deviation.

x = 19.3

$\sigma$ = 11.1

n = 444

significance level, $\alpha$ = 0.05 (If no value is given, we take level of 0.05)

Test statistic $z* = \frac{x-\mu}{\sigma/\sqrt{n}}$

$= \frac{19.3-20}{11.1/\sqrt{444}}$

= - 1.33

The attached files contains additional information

6 0

3 years ago